• Vamos resolver o item \(1\). Os passos são os mesmos para todos os itens; só mudam os valores.

• A equação é

  \[x^{2} + 2 x + y^{2} + 4 y - 4\]

• Vamos separar os termos em \(x\):

  \[x^{2} + 2 x\]

• Completando o quadrado, isto é igual a:

  \[\left(x + 1\right)^{2} - 1\]

• Vamos separar os termos em \(y\):

  \[y^{2} + 4 y\]

• Completando o quadrado, isto é igual a:

  \[\left(y + 2\right)^{2} - 4\]

• Agora, somamos estas duas expressões (mais o termo independente da equação original).

  O resultado é uma equação equivalente à original:

  \[\left(x + 1\right)^{2} + \left(y + 2\right)^{2} - 9 = 0\]

• Jogando o termo independente para o lado direito, chegamos à forma canônica da equação do círculo:

  \[\left(x + 1\right)^{2} + \left(y + 2\right)^{2} = 9\]

As respostas para todos os itens são:

\[ \begin{array}{r|lll} & \textbf{Geral} & \qquad & \textbf{Canônica} \\ \hline 1 & x^{2} + 2 x + y^{2} + 4 y - 4 = 0 & & \left(x + 1\right)^{2} + \left(y + 2\right)^{2} = 9 \\ 2 & x^{2} + 2 x + y^{2} + 4 y + 4 = 0 & & \left(x + 1\right)^{2} + \left(y + 2\right)^{2} = 1 \\ 3 & x^{2} - 4 x + y^{2} - 4 y - 1 = 0 & & \left(x - 2\right)^{2} + \left(y - 2\right)^{2} = 9 \\ 4 & x^{2} - 2 x + y^{2} + 2 y + 1 = 0 & & \left(x - 1\right)^{2} + \left(y + 1\right)^{2} = 1 \\ 5 & x^{2} - 4 x + y^{2} - 2 y + 1 = 0 & & \left(x - 2\right)^{2} + \left(y - 1\right)^{2} = 4 \\ 6 & x^{2} - 2 x + y^{2} - 2 y - 2 = 0 & & \left(x - 1\right)^{2} + \left(y - 1\right)^{2} = 4 \\ 7 & x^{2} + 4 x + y^{2} + 4 y - 1 = 0 & & \left(x + 2\right)^{2} + \left(y + 2\right)^{2} = 9 \\ 8 & x^{2} - 4 x + y^{2} - 4 y - 8 = 0 & & \left(x - 2\right)^{2} + \left(y - 2\right)^{2} = 16 \\ 9 & x^{2} + 4 x + y^{2} - 4 y + 7 = 0 & & \left(x + 2\right)^{2} + \left(y - 2\right)^{2} = 1 \\ 10 & x^{2} + 2 x + y^{2} - 4 y - 4 = 0 & & \left(x + 1\right)^{2} + \left(y - 2\right)^{2} = 9 \\ 11 & x^{2} - 4 x + y^{2} - 2 y - 11 = 0 & & \left(x - 2\right)^{2} + \left(y - 1\right)^{2} = 16 \\ 12 & x^{2} - 4 x + y^{2} + 4 y + 7 = 0 & & \left(x - 2\right)^{2} + \left(y + 2\right)^{2} = 1 \\ 13 & x^{2} + 4 x + y^{2} - 2 y + 1 = 0 & & \left(x + 2\right)^{2} + \left(y - 1\right)^{2} = 4 \\ 14 & x^{2} - 2 x + y^{2} - 2 y - 7 = 0 & & \left(x - 1\right)^{2} + \left(y - 1\right)^{2} = 9 \\ 15 & x^{2} + 2 x + y^{2} - 2 y - 7 = 0 & & \left(x + 1\right)^{2} + \left(y - 1\right)^{2} = 9 \\ 16 & x^{2} - 2 x + y^{2} + 4 y + 1 = 0 & & \left(x - 1\right)^{2} + \left(y + 2\right)^{2} = 4 \\ 17 & x^{2} - 2 x + y^{2} - 2 y - 14 = 0 & & \left(x - 1\right)^{2} + \left(y - 1\right)^{2} = 16 \\ 18 & x^{2} - 2 x + y^{2} + 4 y - 4 = 0 & & \left(x - 1\right)^{2} + \left(y + 2\right)^{2} = 9 \\ 19 & x^{2} - 2 x + y^{2} - 4 y + 4 = 0 & & \left(x - 1\right)^{2} + \left(y - 2\right)^{2} = 1 \\ 20 & x^{2} + 2 x + y^{2} + 2 y - 2 = 0 & & \left(x + 1\right)^{2} + \left(y + 1\right)^{2} = 4 \\ 21 & x^{2} - 4 x + y^{2} - 2 y - 4 = 0 & & \left(x - 2\right)^{2} + \left(y - 1\right)^{2} = 9 \\ 22 & x^{2} + 2 x + y^{2} + 2 y - 7 = 0 & & \left(x + 1\right)^{2} + \left(y + 1\right)^{2} = 9 \\ 23 & x^{2} - 4 x + y^{2} + 2 y - 4 = 0 & & \left(x - 2\right)^{2} + \left(y + 1\right)^{2} = 9 \\ 24 & x^{2} + 2 x + y^{2} - 2 y - 2 = 0 & & \left(x + 1\right)^{2} + \left(y - 1\right)^{2} = 4 \\ 25 & x^{2} + 4 x + y^{2} + 2 y - 4 = 0 & & \left(x + 2\right)^{2} + \left(y + 1\right)^{2} = 9 \\ 26 & x^{2} + 4 x + y^{2} - 4 y - 1 = 0 & & \left(x + 2\right)^{2} + \left(y - 2\right)^{2} = 9 \\ 27 & x^{2} - 4 x + y^{2} - 4 y + 7 = 0 & & \left(x - 2\right)^{2} + \left(y - 2\right)^{2} = 1 \\ 28 & x^{2} - 4 x + y^{2} + 4 y - 1 = 0 & & \left(x - 2\right)^{2} + \left(y + 2\right)^{2} = 9 \\ 29 & x^{2} + 4 x + y^{2} - 2 y - 11 = 0 & & \left(x + 2\right)^{2} + \left(y - 1\right)^{2} = 16 \\ 30 & x^{2} - 2 x + y^{2} - 2 y + 1 = 0 & & \left(x - 1\right)^{2} + \left(y - 1\right)^{2} = 1 \\ 31 & x^{2} + 2 x + y^{2} + 2 y - 14 = 0 & & \left(x + 1\right)^{2} + \left(y + 1\right)^{2} = 16 \\ 32 & x^{2} + 4 x + y^{2} + 2 y + 4 = 0 & & \left(x + 2\right)^{2} + \left(y + 1\right)^{2} = 1 \\ 33 & x^{2} + 4 x + y^{2} - 2 y - 4 = 0 & & \left(x + 2\right)^{2} + \left(y - 1\right)^{2} = 9 \\ 34 & x^{2} + 4 x + y^{2} + 4 y - 8 = 0 & & \left(x + 2\right)^{2} + \left(y + 2\right)^{2} = 16 \\ 35 & x^{2} + 2 x + y^{2} - 2 y + 1 = 0 & & \left(x + 1\right)^{2} + \left(y - 1\right)^{2} = 1 \\ 36 & x^{2} + 2 x + y^{2} - 4 y - 11 = 0 & & \left(x + 1\right)^{2} + \left(y - 2\right)^{2} = 16 \\ 37 & x^{2} - 4 x + y^{2} - 4 y + 4 = 0 & & \left(x - 2\right)^{2} + \left(y - 2\right)^{2} = 4 \\ 38 & x^{2} - 4 x + y^{2} + 2 y + 1 = 0 & & \left(x - 2\right)^{2} + \left(y + 1\right)^{2} = 4 \\ 39 & x^{2} - 4 x + y^{2} + 4 y - 8 = 0 & & \left(x - 2\right)^{2} + \left(y + 2\right)^{2} = 16 \\ 40 & x^{2} - 2 x + y^{2} + 2 y - 7 = 0 & & \left(x - 1\right)^{2} + \left(y + 1\right)^{2} = 9 \\ 41 & x^{2} - 2 x + y^{2} + 2 y - 14 = 0 & & \left(x - 1\right)^{2} + \left(y + 1\right)^{2} = 16 \\ 42 & x^{2} - 2 x + y^{2} - 4 y - 4 = 0 & & \left(x - 1\right)^{2} + \left(y - 2\right)^{2} = 9 \\ \hline \end{array} \]

• Você percebeu que o centro da elipse do nariz é exatamente o centro do círculo do rosto?

• Vamos resolver o item \(1\). Os passos são os mesmos para todos os itens; só mudam os valores.

• A equação canônica de uma elipse com eixo maior horizontal é da forma

  \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]

  onde \((h, k)\) é o centro, que foi dado.

• O valor de \(a\) foi dado.

• O valor de \(b\) pode ser calculado como \(b = \sqrt{a^2 - c^2}\), com \(c\) dado.

• Usando os valores do item \(1\):

  • Centro \({} = \left( -1; \ -2\right)\)

  • \(a = \frac{33}{100}\)

  • \(c = \frac{3}{10}\)

  • Calculamos \(b = \frac{3 \sqrt{21}}{100}\)

  • E a equação fica

    \[\frac{\left(x + 1\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y + 2\right)^{2}}{\frac{189}{10000}} = 1\]

As respostas para todos os itens são:

\[ \begin{array}{r|l} \hline 1 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y + 2\right)^{2}}{\frac{189}{10000}} = 1 \\ 2 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{10000}} + \frac{\left(y + 2\right)^{2}}{\frac{21}{10000}} = 1 \\ 3 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y - 2\right)^{2}}{\frac{189}{10000}} = 1 \\ 4 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{10000}} + \frac{\left(y + 1\right)^{2}}{\frac{21}{10000}} = 1 \\ 5 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{2500}} + \frac{\left(y - 1\right)^{2}}{\frac{21}{2500}} = 1 \\ 6 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{2500}} + \frac{\left(y - 1\right)^{2}}{\frac{21}{2500}} = 1 \\ 7 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y + 2\right)^{2}}{\frac{189}{10000}} = 1 \\ 8 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{625}} + \frac{\left(y - 2\right)^{2}}{\frac{21}{625}} = 1 \\ 9 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{121}{10000}} + \frac{\left(y - 2\right)^{2}}{\frac{21}{10000}} = 1 \\ 10 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y - 2\right)^{2}}{\frac{189}{10000}} = 1 \\ 11 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{625}} + \frac{\left(y - 1\right)^{2}}{\frac{21}{625}} = 1 \\ 12 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{10000}} + \frac{\left(y + 2\right)^{2}}{\frac{21}{10000}} = 1 \\ 13 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{121}{2500}} + \frac{\left(y - 1\right)^{2}}{\frac{21}{2500}} = 1 \\ 14 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y - 1\right)^{2}}{\frac{189}{10000}} = 1 \\ 15 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y - 1\right)^{2}}{\frac{189}{10000}} = 1 \\ 16 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{2500}} + \frac{\left(y + 2\right)^{2}}{\frac{21}{2500}} = 1 \\ 17 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{625}} + \frac{\left(y - 1\right)^{2}}{\frac{21}{625}} = 1 \\ 18 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y + 2\right)^{2}}{\frac{189}{10000}} = 1 \\ 19 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{10000}} + \frac{\left(y - 2\right)^{2}}{\frac{21}{10000}} = 1 \\ 20 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{2500}} + \frac{\left(y + 1\right)^{2}}{\frac{21}{2500}} = 1 \\ 21 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y - 1\right)^{2}}{\frac{189}{10000}} = 1 \\ 22 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y + 1\right)^{2}}{\frac{189}{10000}} = 1 \\ 23 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y + 1\right)^{2}}{\frac{189}{10000}} = 1 \\ 24 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{2500}} + \frac{\left(y - 1\right)^{2}}{\frac{21}{2500}} = 1 \\ 25 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y + 1\right)^{2}}{\frac{189}{10000}} = 1 \\ 26 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y - 2\right)^{2}}{\frac{189}{10000}} = 1 \\ 27 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{10000}} + \frac{\left(y - 2\right)^{2}}{\frac{21}{10000}} = 1 \\ 28 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y + 2\right)^{2}}{\frac{189}{10000}} = 1 \\ 29 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{121}{625}} + \frac{\left(y - 1\right)^{2}}{\frac{21}{625}} = 1 \\ 30 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{10000}} + \frac{\left(y - 1\right)^{2}}{\frac{21}{10000}} = 1 \\ 31 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{625}} + \frac{\left(y + 1\right)^{2}}{\frac{21}{625}} = 1 \\ 32 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{121}{10000}} + \frac{\left(y + 1\right)^{2}}{\frac{21}{10000}} = 1 \\ 33 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y - 1\right)^{2}}{\frac{189}{10000}} = 1 \\ 34 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{121}{625}} + \frac{\left(y + 2\right)^{2}}{\frac{21}{625}} = 1 \\ 35 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{10000}} + \frac{\left(y - 1\right)^{2}}{\frac{21}{10000}} = 1 \\ 36 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{625}} + \frac{\left(y - 2\right)^{2}}{\frac{21}{625}} = 1 \\ 37 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{2500}} + \frac{\left(y - 2\right)^{2}}{\frac{21}{2500}} = 1 \\ 38 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{2500}} + \frac{\left(y + 1\right)^{2}}{\frac{21}{2500}} = 1 \\ 39 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{625}} + \frac{\left(y + 2\right)^{2}}{\frac{21}{625}} = 1 \\ 40 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y + 1\right)^{2}}{\frac{189}{10000}} = 1 \\ 41 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{625}} + \frac{\left(y + 1\right)^{2}}{\frac{21}{625}} = 1 \\ 42 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{1089}{10000}} + \frac{\left(y - 2\right)^{2}}{\frac{189}{10000}} = 1 \\ \hline \end{array} \]

• Vamos resolver o item \(1\). Os passos são os mesmos para todos os itens; só mudam os valores.

• A equação é

  \[x^{2} + 2 x + \frac{121 y^{2}}{21} + \frac{242 y}{15} + \frac{2297}{240} = 0\]

• Vamos separar os termos em \(x\):

  \[x^{2} + 2 x\]

• Completando o quadrado, isto é igual a:

  \[\left(x + 1\right)^{2} - 1\]

• Vamos separar os termos em \(y\):

  \[\frac{121 y^{2}}{21} + \frac{242 y}{15}\]

• Completando o quadrado, isto é igual a:

  \[\frac{121 \left(y + \frac{7}{5}\right)^{2}}{21} - \frac{847}{75}\]

• Agora, somamos estas duas expressões (mais o termo independente da equação original).

  O resultado é uma equação equivalente à original:

  \[\left(x + 1\right)^{2} + \frac{121 \left(y + \frac{7}{5}\right)^{2}}{21} - \frac{1089}{400} = 0\]

• Jogando o termo independente para o lado direito, e dividindo tudo para o lado direito ficar igual a \(1\), chegamos à forma canônica da equação da elipse:

  \[\frac{\left(x + 1\right)^{2}}{\frac{1089}{400}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{\frac{189}{400}} = 1\]

As respostas para todos os itens são:

\[ \begin{array}{r|l}\hline 1 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{1089}{400}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 2 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{400}} + \frac{\left(y + \frac{9}{5}\right)^{2}}{\frac{21}{400}} = 1 \\ 3 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{1089}{400}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 4 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{400}} + \frac{\left(y + \frac{4}{5}\right)^{2}}{\frac{21}{400}} = 1 \\ 5 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{100}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{21}{100}} = 1 \\ 6 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{100}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{21}{100}} = 1 \\ 7 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{1089}{400}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 8 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{25}} + \frac{\left(y - \frac{14}{5}\right)^{2}}{\frac{21}{25}} = 1 \\ 9 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{121}{400}} + \frac{\left(y - \frac{11}{5}\right)^{2}}{\frac{21}{400}} = 1 \\ 10 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{1089}{400}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 11 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{25}} + \frac{\left(y - \frac{9}{5}\right)^{2}}{\frac{21}{25}} = 1 \\ 12 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{400}} + \frac{\left(y + \frac{9}{5}\right)^{2}}{\frac{21}{400}} = 1 \\ 13 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{121}{100}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{21}{100}} = 1 \\ 14 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{1089}{400}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 15 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{1089}{400}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 16 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{100}} + \frac{\left(y + \frac{8}{5}\right)^{2}}{\frac{21}{100}} = 1 \\ 17 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{25}} + \frac{\left(y - \frac{9}{5}\right)^{2}}{\frac{21}{25}} = 1 \\ 18 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{1089}{400}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 19 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{400}} + \frac{\left(y - \frac{11}{5}\right)^{2}}{\frac{21}{400}} = 1 \\ 20 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{100}} + \frac{\left(y + \frac{3}{5}\right)^{2}}{\frac{21}{100}} = 1 \\ 21 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{1089}{400}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 22 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{1089}{400}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 23 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{1089}{400}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 24 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{100}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{21}{100}} = 1 \\ 25 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{1089}{400}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 26 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{1089}{400}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 27 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{400}} + \frac{\left(y - \frac{11}{5}\right)^{2}}{\frac{21}{400}} = 1 \\ 28 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{1089}{400}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 29 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{121}{25}} + \frac{\left(y - \frac{9}{5}\right)^{2}}{\frac{21}{25}} = 1 \\ 30 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{400}} + \frac{\left(y - \frac{6}{5}\right)^{2}}{\frac{21}{400}} = 1 \\ 31 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{25}} + \frac{\left(y + \frac{1}{5}\right)^{2}}{\frac{21}{25}} = 1 \\ 32 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{121}{400}} + \frac{\left(y + \frac{4}{5}\right)^{2}}{\frac{21}{400}} = 1 \\ 33 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{1089}{400}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 34 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{121}{25}} + \frac{\left(y + \frac{6}{5}\right)^{2}}{\frac{21}{25}} = 1 \\ 35 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{400}} + \frac{\left(y - \frac{6}{5}\right)^{2}}{\frac{21}{400}} = 1 \\ 36 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{121}{25}} + \frac{\left(y - \frac{14}{5}\right)^{2}}{\frac{21}{25}} = 1 \\ 37 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{100}} + \frac{\left(y - \frac{12}{5}\right)^{2}}{\frac{21}{100}} = 1 \\ 38 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{100}} + \frac{\left(y + \frac{3}{5}\right)^{2}}{\frac{21}{100}} = 1 \\ 39 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{121}{25}} + \frac{\left(y + \frac{6}{5}\right)^{2}}{\frac{21}{25}} = 1 \\ 40 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{1089}{400}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ 41 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{121}{25}} + \frac{\left(y + \frac{1}{5}\right)^{2}}{\frac{21}{25}} = 1 \\ 42 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{1089}{400}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{\frac{189}{400}} = 1 \\ \hline \end{array} \]

• Para achar os focos, lembre-se de que, para uma elipse com eixo maior horizontal, com centro em \((h, k)\), os focos são

  \[ \begin{align} F_1 &= (h - c, k) \\ F_2 &= (h + c, k) \end{align} \]

  onde \(c\) é a distância focal, que pode ser calculada como

  \[ c = \sqrt{a^2 - b^2} \]

• Para o item \(1\)

  \[ \begin{align} h &= -1 \\ k &= -7/5 \\ c &= \sqrt{1089/400 - 189/400} = 3/2 \end{align} \]

• Os focos são

  \[ \left( - \frac{5}{2}; \ - \frac{7}{5}\right)\quad\text{e}\quad\left( \frac{1}{2}; \ - \frac{7}{5}\right) \]

As respostas para todos os itens são:

\[ \begin{array}{r|ll}\hline 1 & \displaystyle \left( - \frac{5}{2}; \ - \frac{7}{5}\right) & \displaystyle \left( \frac{1}{2}; \ - \frac{7}{5}\right) \\ 2 & \displaystyle \left( - \frac{3}{2}; \ - \frac{9}{5}\right) & \displaystyle \left( - \frac{1}{2}; \ - \frac{9}{5}\right) \\ 3 & \displaystyle \left( \frac{1}{2}; \ \frac{13}{5}\right) & \displaystyle \left( \frac{7}{2}; \ \frac{13}{5}\right) \\ 4 & \displaystyle \left( \frac{1}{2}; \ - \frac{4}{5}\right) & \displaystyle \left( \frac{3}{2}; \ - \frac{4}{5}\right) \\ 5 & \displaystyle \left( 1; \ \frac{7}{5}\right) & \displaystyle \left( 3; \ \frac{7}{5}\right) \\ 6 & \displaystyle \left( 0; \ \frac{7}{5}\right) & \displaystyle \left( 2; \ \frac{7}{5}\right) \\ 7 & \displaystyle \left( - \frac{7}{2}; \ - \frac{7}{5}\right) & \displaystyle \left( - \frac{1}{2}; \ - \frac{7}{5}\right) \\ 8 & \displaystyle \left( 0; \ \frac{14}{5}\right) & \displaystyle \left( 4; \ \frac{14}{5}\right) \\ 9 & \displaystyle \left( - \frac{5}{2}; \ \frac{11}{5}\right) & \displaystyle \left( - \frac{3}{2}; \ \frac{11}{5}\right) \\ 10 & \displaystyle \left( - \frac{5}{2}; \ \frac{13}{5}\right) & \displaystyle \left( \frac{1}{2}; \ \frac{13}{5}\right) \\ 11 & \displaystyle \left( 0; \ \frac{9}{5}\right) & \displaystyle \left( 4; \ \frac{9}{5}\right) \\ 12 & \displaystyle \left( \frac{3}{2}; \ - \frac{9}{5}\right) & \displaystyle \left( \frac{5}{2}; \ - \frac{9}{5}\right) \\ 13 & \displaystyle \left( -3; \ \frac{7}{5}\right) & \displaystyle \left( -1; \ \frac{7}{5}\right) \\ 14 & \displaystyle \left( - \frac{1}{2}; \ \frac{8}{5}\right) & \displaystyle \left( \frac{5}{2}; \ \frac{8}{5}\right) \\ 15 & \displaystyle \left( - \frac{5}{2}; \ \frac{8}{5}\right) & \displaystyle \left( \frac{1}{2}; \ \frac{8}{5}\right) \\ 16 & \displaystyle \left( 0; \ - \frac{8}{5}\right) & \displaystyle \left( 2; \ - \frac{8}{5}\right) \\ 17 & \displaystyle \left( -1; \ \frac{9}{5}\right) & \displaystyle \left( 3; \ \frac{9}{5}\right) \\ 18 & \displaystyle \left( - \frac{1}{2}; \ - \frac{7}{5}\right) & \displaystyle \left( \frac{5}{2}; \ - \frac{7}{5}\right) \\ 19 & \displaystyle \left( \frac{1}{2}; \ \frac{11}{5}\right) & \displaystyle \left( \frac{3}{2}; \ \frac{11}{5}\right) \\ 20 & \displaystyle \left( -2; \ - \frac{3}{5}\right) & \displaystyle \left( 0; \ - \frac{3}{5}\right) \\ 21 & \displaystyle \left( \frac{1}{2}; \ \frac{8}{5}\right) & \displaystyle \left( \frac{7}{2}; \ \frac{8}{5}\right) \\ 22 & \displaystyle \left( - \frac{5}{2}; \ - \frac{2}{5}\right) & \displaystyle \left( \frac{1}{2}; \ - \frac{2}{5}\right) \\ 23 & \displaystyle \left( \frac{1}{2}; \ - \frac{2}{5}\right) & \displaystyle \left( \frac{7}{2}; \ - \frac{2}{5}\right) \\ 24 & \displaystyle \left( -2; \ \frac{7}{5}\right) & \displaystyle \left( 0; \ \frac{7}{5}\right) \\ 25 & \displaystyle \left( - \frac{7}{2}; \ - \frac{2}{5}\right) & \displaystyle \left( - \frac{1}{2}; \ - \frac{2}{5}\right) \\ 26 & \displaystyle \left( - \frac{7}{2}; \ \frac{13}{5}\right) & \displaystyle \left( - \frac{1}{2}; \ \frac{13}{5}\right) \\ 27 & \displaystyle \left( \frac{3}{2}; \ \frac{11}{5}\right) & \displaystyle \left( \frac{5}{2}; \ \frac{11}{5}\right) \\ 28 & \displaystyle \left( \frac{1}{2}; \ - \frac{7}{5}\right) & \displaystyle \left( \frac{7}{2}; \ - \frac{7}{5}\right) \\ 29 & \displaystyle \left( -4; \ \frac{9}{5}\right) & \displaystyle \left( 0; \ \frac{9}{5}\right) \\ 30 & \displaystyle \left( \frac{1}{2}; \ \frac{6}{5}\right) & \displaystyle \left( \frac{3}{2}; \ \frac{6}{5}\right) \\ 31 & \displaystyle \left( -3; \ - \frac{1}{5}\right) & \displaystyle \left( 1; \ - \frac{1}{5}\right) \\ 32 & \displaystyle \left( - \frac{5}{2}; \ - \frac{4}{5}\right) & \displaystyle \left( - \frac{3}{2}; \ - \frac{4}{5}\right) \\ 33 & \displaystyle \left( - \frac{7}{2}; \ \frac{8}{5}\right) & \displaystyle \left( - \frac{1}{2}; \ \frac{8}{5}\right) \\ 34 & \displaystyle \left( -4; \ - \frac{6}{5}\right) & \displaystyle \left( 0; \ - \frac{6}{5}\right) \\ 35 & \displaystyle \left( - \frac{3}{2}; \ \frac{6}{5}\right) & \displaystyle \left( - \frac{1}{2}; \ \frac{6}{5}\right) \\ 36 & \displaystyle \left( -3; \ \frac{14}{5}\right) & \displaystyle \left( 1; \ \frac{14}{5}\right) \\ 37 & \displaystyle \left( 1; \ \frac{12}{5}\right) & \displaystyle \left( 3; \ \frac{12}{5}\right) \\ 38 & \displaystyle \left( 1; \ - \frac{3}{5}\right) & \displaystyle \left( 3; \ - \frac{3}{5}\right) \\ 39 & \displaystyle \left( 0; \ - \frac{6}{5}\right) & \displaystyle \left( 4; \ - \frac{6}{5}\right) \\ 40 & \displaystyle \left( - \frac{1}{2}; \ - \frac{2}{5}\right) & \displaystyle \left( \frac{5}{2}; \ - \frac{2}{5}\right) \\ 41 & \displaystyle \left( -1; \ - \frac{1}{5}\right) & \displaystyle \left( 3; \ - \frac{1}{5}\right) \\ 42 & \displaystyle \left( - \frac{1}{2}; \ \frac{13}{5}\right) & \displaystyle \left( \frac{5}{2}; \ \frac{13}{5}\right) \\ \hline \end{array} \]

• Vamos resolver o item \(1\). Os passos são os mesmos para todos os itens; só mudam os valores.

• Temos, para cada olho:

  • O centro \((h, k)\), que é um dos focos da elipse do item anterior;

  • O valor de \(a\);

  • O valor de \(c\).

• O mais prático é achar a equação canônica de cada olho, que é da forma

  \[ \frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1 \]

  onde \(b = \sqrt{a^2 - c^2}\).

• Importante: como a elipse tem o eixo maior vertical, o denominador \(a^2\) aparece no termo em \(y\).

• Uma vez achada a equação canônica, desenvolvemos os quadrados e a soma para achar a equação geral.

• Para um olho:

  • Centro \((h, k) = \left( - \frac{5}{2}; \ - \frac{7}{5}\right)\)

  • \(a = 1\)

  • \(c = \frac{3}{4}\)

  • \(b = \frac{\sqrt{7}}{4}\)

  • Equação canônica \({}= \frac{\left(x + \frac{5}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{1} = 1\)

  • Equação geral \({}= \frac{16 x^{2}}{7} + \frac{80 x}{7} + y^{2} + \frac{14 y}{5} + \frac{2668}{175} = 0\)

• Para o outro olho:

  • Centro \((h, k) = \left( \frac{1}{2}; \ - \frac{7}{5}\right)\)

  • \(a = 1\)

  • \(c = \frac{3}{4}\)

  • \(b = \frac{\sqrt{7}}{4}\)

  • Equação canônica \({}= \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{1} = 1\)

  • Equação geral \({}= \frac{16 x^{2}}{7} - \frac{16 x}{7} + y^{2} + \frac{14 y}{5} + \frac{268}{175} = 0\)

As respostas para todos os itens (equações canônicas e gerais do primeiro olho) são:

\[ \begin{array}{r|lll} \hline 1 & \displaystyle \frac{\left(x + \frac{5}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{80 x}{7} + y^{2} + \frac{14 y}{5} + \frac{2668}{175} = 0 \\ 2 & \displaystyle \frac{\left(x + \frac{3}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y + \frac{9}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} + \frac{432 x}{7} + 9 y^{2} + \frac{162 y}{5} + \frac{13028}{175} = 0 \\ 3 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{16 x}{7} + y^{2} - \frac{26 y}{5} + \frac{1108}{175} = 0 \\ 4 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y + \frac{4}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} - \frac{144 x}{7} + 9 y^{2} + \frac{72 y}{5} + \frac{1733}{175} = 0 \\ 5 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{7}{36}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} - \frac{72 x}{7} + \frac{9 y^{2}}{4} - \frac{63 y}{10} + \frac{5987}{700} = 0 \\ 6 & \displaystyle \frac{x^{2}}{\frac{7}{36}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} + \frac{9 y^{2}}{4} - \frac{63 y}{10} + \frac{341}{100} = 0 \\ 7 & \displaystyle \frac{\left(x + \frac{7}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + 16 x + y^{2} + \frac{14 y}{5} + \frac{724}{25} = 0 \\ 8 & \displaystyle \frac{x^{2}}{\frac{7}{9}} + \frac{\left(y - \frac{14}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{9 y^{2}}{16} - \frac{63 y}{20} + \frac{341}{100} = 0 \\ 9 & \displaystyle \frac{\left(x + \frac{5}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y - \frac{11}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} + \frac{720 x}{7} + 9 y^{2} - \frac{198 y}{5} + \frac{29948}{175} = 0 \\ 10 & \displaystyle \frac{\left(x + \frac{5}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{80 x}{7} + y^{2} - \frac{26 y}{5} + \frac{3508}{175} = 0 \\ 11 & \displaystyle \frac{x^{2}}{\frac{7}{9}} + \frac{\left(y - \frac{9}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{9 y^{2}}{16} - \frac{81 y}{40} + \frac{329}{400} = 0 \\ 12 & \displaystyle \frac{\left(x - \frac{3}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y + \frac{9}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} - \frac{432 x}{7} + 9 y^{2} + \frac{162 y}{5} + \frac{13028}{175} = 0 \\ 13 & \displaystyle \frac{\left(x + 3\right)^{2}}{\frac{7}{36}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} + \frac{216 x}{7} + \frac{9 y^{2}}{4} - \frac{63 y}{10} + \frac{34787}{700} = 0 \\ 14 & \displaystyle \frac{\left(x + \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{16 x}{7} + y^{2} - \frac{16 y}{5} + \frac{373}{175} = 0 \\ 15 & \displaystyle \frac{\left(x + \frac{5}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{80 x}{7} + y^{2} - \frac{16 y}{5} + \frac{2773}{175} = 0 \\ 16 & \displaystyle \frac{x^{2}}{\frac{7}{36}} + \frac{\left(y + \frac{8}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} + \frac{9 y^{2}}{4} + \frac{36 y}{5} + \frac{119}{25} = 0 \\ 17 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{7}{9}} + \frac{\left(y - \frac{9}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{18 x}{7} + \frac{9 y^{2}}{16} - \frac{81 y}{40} + \frac{5903}{2800} = 0 \\ 18 & \displaystyle \frac{\left(x + \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{16 x}{7} + y^{2} + \frac{14 y}{5} + \frac{268}{175} = 0 \\ 19 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y - \frac{11}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} - \frac{144 x}{7} + 9 y^{2} - \frac{198 y}{5} + \frac{8348}{175} = 0 \\ 20 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{7}{36}} + \frac{\left(y + \frac{3}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} + \frac{144 x}{7} + \frac{9 y^{2}}{4} + \frac{27 y}{10} + \frac{14267}{700} = 0 \\ 21 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{16 x}{7} + y^{2} - \frac{16 y}{5} + \frac{373}{175} = 0 \\ 22 & \displaystyle \frac{\left(x + \frac{5}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{80 x}{7} + y^{2} + \frac{4 y}{5} + \frac{2353}{175} = 0 \\ 23 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{16 x}{7} + y^{2} + \frac{4 y}{5} - \frac{47}{175} = 0 \\ 24 & \displaystyle \frac{\left(x + 2\right)^{2}}{\frac{7}{36}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} + \frac{144 x}{7} + \frac{9 y^{2}}{4} - \frac{63 y}{10} + \frac{16787}{700} = 0 \\ 25 & \displaystyle \frac{\left(x + \frac{7}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + 16 x + y^{2} + \frac{4 y}{5} + \frac{679}{25} = 0 \\ 26 & \displaystyle \frac{\left(x + \frac{7}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + 16 x + y^{2} - \frac{26 y}{5} + \frac{844}{25} = 0 \\ 27 & \displaystyle \frac{\left(x - \frac{3}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y - \frac{11}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} - \frac{432 x}{7} + 9 y^{2} - \frac{198 y}{5} + \frac{15548}{175} = 0 \\ 28 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{16 x}{7} + y^{2} + \frac{14 y}{5} + \frac{268}{175} = 0 \\ 29 & \displaystyle \frac{\left(x + 4\right)^{2}}{\frac{7}{9}} + \frac{\left(y - \frac{9}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{72 x}{7} + \frac{9 y^{2}}{16} - \frac{81 y}{40} + \frac{59903}{2800} = 0 \\ 30 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y - \frac{6}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} - \frac{144 x}{7} + 9 y^{2} - \frac{108 y}{5} + \frac{2993}{175} = 0 \\ 31 & \displaystyle \frac{\left(x + 3\right)^{2}}{\frac{7}{9}} + \frac{\left(y + \frac{1}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{54 x}{7} + \frac{9 y^{2}}{16} + \frac{9 y}{40} + \frac{29663}{2800} = 0 \\ 32 & \displaystyle \frac{\left(x + \frac{5}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y + \frac{4}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} + \frac{720 x}{7} + 9 y^{2} + \frac{72 y}{5} + \frac{23333}{175} = 0 \\ 33 & \displaystyle \frac{\left(x + \frac{7}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + 16 x + y^{2} - \frac{16 y}{5} + \frac{739}{25} = 0 \\ 34 & \displaystyle \frac{\left(x + 4\right)^{2}}{\frac{7}{9}} + \frac{\left(y + \frac{6}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{72 x}{7} + \frac{9 y^{2}}{16} + \frac{27 y}{20} + \frac{14267}{700} = 0 \\ 35 & \displaystyle \frac{\left(x + \frac{3}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y - \frac{6}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} + \frac{432 x}{7} + 9 y^{2} - \frac{108 y}{5} + \frac{10193}{175} = 0 \\ 36 & \displaystyle \frac{\left(x + 3\right)^{2}}{\frac{7}{9}} + \frac{\left(y - \frac{14}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{54 x}{7} + \frac{9 y^{2}}{16} - \frac{63 y}{20} + \frac{10487}{700} = 0 \\ 37 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{7}{36}} + \frac{\left(y - \frac{12}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} - \frac{72 x}{7} + \frac{9 y^{2}}{4} - \frac{54 y}{5} + \frac{2993}{175} = 0 \\ 38 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{7}{36}} + \frac{\left(y + \frac{3}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} - \frac{72 x}{7} + \frac{9 y^{2}}{4} + \frac{27 y}{10} + \frac{3467}{700} = 0 \\ 39 & \displaystyle \frac{x^{2}}{\frac{7}{9}} + \frac{\left(y + \frac{6}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{9 y^{2}}{16} + \frac{27 y}{20} - \frac{19}{100} = 0 \\ 40 & \displaystyle \frac{\left(x + \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{16 x}{7} + y^{2} + \frac{4 y}{5} - \frac{47}{175} = 0 \\ 41 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{7}{9}} + \frac{\left(y + \frac{1}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{18 x}{7} + \frac{9 y^{2}}{16} + \frac{9 y}{40} + \frac{863}{2800} = 0 \\ 42 & \displaystyle \frac{\left(x + \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{16 x}{7} + y^{2} - \frac{26 y}{5} + \frac{1108}{175} = 0 \\ \hline \end{array} \]

As respostas para todos os itens (equações canônicas e gerais do segundo olho) são:

\[ \begin{array}{r|lll} \hline 1 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{16 x}{7} + y^{2} + \frac{14 y}{5} + \frac{268}{175} = 0 \\ 2 & \displaystyle \frac{\left(x + \frac{1}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y + \frac{9}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} + \frac{144 x}{7} + 9 y^{2} + \frac{162 y}{5} + \frac{5828}{175} = 0 \\ 3 & \displaystyle \frac{\left(x - \frac{7}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - 16 x + y^{2} - \frac{26 y}{5} + \frac{844}{25} = 0 \\ 4 & \displaystyle \frac{\left(x - \frac{3}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y + \frac{4}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} - \frac{432 x}{7} + 9 y^{2} + \frac{72 y}{5} + \frac{8933}{175} = 0 \\ 5 & \displaystyle \frac{\left(x - 3\right)^{2}}{\frac{7}{36}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} - \frac{216 x}{7} + \frac{9 y^{2}}{4} - \frac{63 y}{10} + \frac{34787}{700} = 0 \\ 6 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{7}{36}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} - \frac{144 x}{7} + \frac{9 y^{2}}{4} - \frac{63 y}{10} + \frac{16787}{700} = 0 \\ 7 & \displaystyle \frac{\left(x + \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{16 x}{7} + y^{2} + \frac{14 y}{5} + \frac{268}{175} = 0 \\ 8 & \displaystyle \frac{\left(x - 4\right)^{2}}{\frac{7}{9}} + \frac{\left(y - \frac{14}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} - \frac{72 x}{7} + \frac{9 y^{2}}{16} - \frac{63 y}{20} + \frac{16787}{700} = 0 \\ 9 & \displaystyle \frac{\left(x + \frac{3}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y - \frac{11}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} + \frac{432 x}{7} + 9 y^{2} - \frac{198 y}{5} + \frac{15548}{175} = 0 \\ 10 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{16 x}{7} + y^{2} - \frac{26 y}{5} + \frac{1108}{175} = 0 \\ 11 & \displaystyle \frac{\left(x - 4\right)^{2}}{\frac{7}{9}} + \frac{\left(y - \frac{9}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} - \frac{72 x}{7} + \frac{9 y^{2}}{16} - \frac{81 y}{40} + \frac{59903}{2800} = 0 \\ 12 & \displaystyle \frac{\left(x - \frac{5}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y + \frac{9}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} - \frac{720 x}{7} + 9 y^{2} + \frac{162 y}{5} + \frac{27428}{175} = 0 \\ 13 & \displaystyle \frac{\left(x + 1\right)^{2}}{\frac{7}{36}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} + \frac{72 x}{7} + \frac{9 y^{2}}{4} - \frac{63 y}{10} + \frac{5987}{700} = 0 \\ 14 & \displaystyle \frac{\left(x - \frac{5}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{80 x}{7} + y^{2} - \frac{16 y}{5} + \frac{2773}{175} = 0 \\ 15 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{16 x}{7} + y^{2} - \frac{16 y}{5} + \frac{373}{175} = 0 \\ 16 & \displaystyle \frac{\left(x - 2\right)^{2}}{\frac{7}{36}} + \frac{\left(y + \frac{8}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} - \frac{144 x}{7} + \frac{9 y^{2}}{4} + \frac{36 y}{5} + \frac{4433}{175} = 0 \\ 17 & \displaystyle \frac{\left(x - 3\right)^{2}}{\frac{7}{9}} + \frac{\left(y - \frac{9}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} - \frac{54 x}{7} + \frac{9 y^{2}}{16} - \frac{81 y}{40} + \frac{34703}{2800} = 0 \\ 18 & \displaystyle \frac{\left(x - \frac{5}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{80 x}{7} + y^{2} + \frac{14 y}{5} + \frac{2668}{175} = 0 \\ 19 & \displaystyle \frac{\left(x - \frac{3}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y - \frac{11}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} - \frac{432 x}{7} + 9 y^{2} - \frac{198 y}{5} + \frac{15548}{175} = 0 \\ 20 & \displaystyle \frac{x^{2}}{\frac{7}{36}} + \frac{\left(y + \frac{3}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} + \frac{9 y^{2}}{4} + \frac{27 y}{10} - \frac{19}{100} = 0 \\ 21 & \displaystyle \frac{\left(x - \frac{7}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - 16 x + y^{2} - \frac{16 y}{5} + \frac{739}{25} = 0 \\ 22 & \displaystyle \frac{\left(x - \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{16 x}{7} + y^{2} + \frac{4 y}{5} - \frac{47}{175} = 0 \\ 23 & \displaystyle \frac{\left(x - \frac{7}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - 16 x + y^{2} + \frac{4 y}{5} + \frac{679}{25} = 0 \\ 24 & \displaystyle \frac{x^{2}}{\frac{7}{36}} + \frac{\left(y - \frac{7}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} + \frac{9 y^{2}}{4} - \frac{63 y}{10} + \frac{341}{100} = 0 \\ 25 & \displaystyle \frac{\left(x + \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{16 x}{7} + y^{2} + \frac{4 y}{5} - \frac{47}{175} = 0 \\ 26 & \displaystyle \frac{\left(x + \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{16 x}{7} + y^{2} - \frac{26 y}{5} + \frac{1108}{175} = 0 \\ 27 & \displaystyle \frac{\left(x - \frac{5}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y - \frac{11}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} - \frac{720 x}{7} + 9 y^{2} - \frac{198 y}{5} + \frac{29948}{175} = 0 \\ 28 & \displaystyle \frac{\left(x - \frac{7}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{7}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - 16 x + y^{2} + \frac{14 y}{5} + \frac{724}{25} = 0 \\ 29 & \displaystyle \frac{x^{2}}{\frac{7}{9}} + \frac{\left(y - \frac{9}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{9 y^{2}}{16} - \frac{81 y}{40} + \frac{329}{400} = 0 \\ 30 & \displaystyle \frac{\left(x - \frac{3}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y - \frac{6}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} - \frac{432 x}{7} + 9 y^{2} - \frac{108 y}{5} + \frac{10193}{175} = 0 \\ 31 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{7}{9}} + \frac{\left(y + \frac{1}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} - \frac{18 x}{7} + \frac{9 y^{2}}{16} + \frac{9 y}{40} + \frac{863}{2800} = 0 \\ 32 & \displaystyle \frac{\left(x + \frac{3}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y + \frac{4}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} + \frac{432 x}{7} + 9 y^{2} + \frac{72 y}{5} + \frac{8933}{175} = 0 \\ 33 & \displaystyle \frac{\left(x + \frac{1}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{8}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} + \frac{16 x}{7} + y^{2} - \frac{16 y}{5} + \frac{373}{175} = 0 \\ 34 & \displaystyle \frac{x^{2}}{\frac{7}{9}} + \frac{\left(y + \frac{6}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} + \frac{9 y^{2}}{16} + \frac{27 y}{20} - \frac{19}{100} = 0 \\ 35 & \displaystyle \frac{\left(x + \frac{1}{2}\right)^{2}}{\frac{7}{144}} + \frac{1}{\frac{1}{9}} \left(y - \frac{6}{5}\right)^{2} = 1 & \qquad & \displaystyle \frac{144 x^{2}}{7} + \frac{144 x}{7} + 9 y^{2} - \frac{108 y}{5} + \frac{2993}{175} = 0 \\ 36 & \displaystyle \frac{\left(x - 1\right)^{2}}{\frac{7}{9}} + \frac{\left(y - \frac{14}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} - \frac{18 x}{7} + \frac{9 y^{2}}{16} - \frac{63 y}{20} + \frac{3287}{700} = 0 \\ 37 & \displaystyle \frac{\left(x - 3\right)^{2}}{\frac{7}{36}} + \frac{\left(y - \frac{12}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} - \frac{216 x}{7} + \frac{9 y^{2}}{4} - \frac{54 y}{5} + \frac{10193}{175} = 0 \\ 38 & \displaystyle \frac{\left(x - 3\right)^{2}}{\frac{7}{36}} + \frac{\left(y + \frac{3}{5}\right)^{2}}{\frac{4}{9}} = 1 & \qquad & \displaystyle \frac{36 x^{2}}{7} - \frac{216 x}{7} + \frac{9 y^{2}}{4} + \frac{27 y}{10} + \frac{32267}{700} = 0 \\ 39 & \displaystyle \frac{\left(x - 4\right)^{2}}{\frac{7}{9}} + \frac{\left(y + \frac{6}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} - \frac{72 x}{7} + \frac{9 y^{2}}{16} + \frac{27 y}{20} + \frac{14267}{700} = 0 \\ 40 & \displaystyle \frac{\left(x - \frac{5}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y + \frac{2}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{80 x}{7} + y^{2} + \frac{4 y}{5} + \frac{2353}{175} = 0 \\ 41 & \displaystyle \frac{\left(x - 3\right)^{2}}{\frac{7}{9}} + \frac{\left(y + \frac{1}{5}\right)^{2}}{\frac{16}{9}} = 1 & \qquad & \displaystyle \frac{9 x^{2}}{7} - \frac{54 x}{7} + \frac{9 y^{2}}{16} + \frac{9 y}{40} + \frac{29663}{2800} = 0 \\ 42 & \displaystyle \frac{\left(x - \frac{5}{2}\right)^{2}}{\frac{7}{16}} + \frac{\left(y - \frac{13}{5}\right)^{2}}{1} = 1 & \qquad & \displaystyle \frac{16 x^{2}}{7} - \frac{80 x}{7} + y^{2} - \frac{26 y}{5} + \frac{3508}{175} = 0 \\ \hline \end{array} \]

• Vamos resolver o item 1. Os passos são os mesmos para todos os itens; só mudam os valores.

• A equação canônica da parábola com eixo vertical é

  \[ y - k = \frac{1}{4p}(x - h)^2 \]

  onde

  • \((h, k)\) é o vértice;

  • \(p = 3/4\) é a distância focal, que foi dada.

• As coordenadas do foco \(F = (-1, -3)\) foram dadas, e o vértice está exatamente a \(p = 3/4\) unidades de distância abaixo do foco.

• Ou seja, \(h = -1\) e \(k = -3 - 3/4 = -15/4\).

• A equação canônica fica

  \[ y + \frac{15}{4} = \frac{1}{3}(x + 1)^2 \]

As respostas para todos os itens são:

\[ \begin{array}{r|l} \hline 1 & \displaystyle y + \frac{15}{4} = \frac{\left(x + 1\right)^{2}}{3} \\ 2 & \displaystyle y + \frac{31}{12} = \left(x + 1\right)^{2} \\ 3 & \displaystyle y - \frac{1}{4} = \frac{\left(x - 2\right)^{2}}{3} \\ 4 & \displaystyle y + \frac{19}{12} = \left(x - 1\right)^{2} \\ 5 & \displaystyle y + \frac{1}{6} = \frac{\left(x - 2\right)^{2}}{2} \\ 6 & \displaystyle y + \frac{1}{6} = \frac{\left(x - 1\right)^{2}}{2} \\ 7 & \displaystyle y + \frac{15}{4} = \frac{\left(x + 2\right)^{2}}{3} \\ 8 & \displaystyle y + \frac{1}{3} = \frac{\left(x - 2\right)^{2}}{4} \\ 9 & \displaystyle y - \frac{17}{12} = \left(x + 2\right)^{2} \\ 10 & \displaystyle y - \frac{1}{4} = \frac{\left(x + 1\right)^{2}}{3} \\ 11 & \displaystyle y + \frac{4}{3} = \frac{\left(x - 2\right)^{2}}{4} \\ 12 & \displaystyle y + \frac{31}{12} = \left(x - 2\right)^{2} \\ 13 & \displaystyle y + \frac{1}{6} = \frac{\left(x + 2\right)^{2}}{2} \\ 14 & \displaystyle y + \frac{3}{4} = \frac{\left(x - 1\right)^{2}}{3} \\ 15 & \displaystyle y + \frac{3}{4} = \frac{\left(x + 1\right)^{2}}{3} \\ 16 & \displaystyle y + \frac{19}{6} = \frac{\left(x - 1\right)^{2}}{2} \\ 17 & \displaystyle y + \frac{4}{3} = \frac{\left(x - 1\right)^{2}}{4} \\ 18 & \displaystyle y + \frac{15}{4} = \frac{\left(x - 1\right)^{2}}{3} \\ 19 & \displaystyle y - \frac{17}{12} = \left(x - 1\right)^{2} \\ 20 & \displaystyle y + \frac{13}{6} = \frac{\left(x + 1\right)^{2}}{2} \\ 21 & \displaystyle y + \frac{3}{4} = \frac{\left(x - 2\right)^{2}}{3} \\ 22 & \displaystyle y + \frac{11}{4} = \frac{\left(x + 1\right)^{2}}{3} \\ 23 & \displaystyle y + \frac{11}{4} = \frac{\left(x - 2\right)^{2}}{3} \\ 24 & \displaystyle y + \frac{1}{6} = \frac{\left(x + 1\right)^{2}}{2} \\ 25 & \displaystyle y + \frac{11}{4} = \frac{\left(x + 2\right)^{2}}{3} \\ 26 & \displaystyle y - \frac{1}{4} = \frac{\left(x + 2\right)^{2}}{3} \\ 27 & \displaystyle y - \frac{17}{12} = \left(x - 2\right)^{2} \\ 28 & \displaystyle y + \frac{15}{4} = \frac{\left(x - 2\right)^{2}}{3} \\ 29 & \displaystyle y + \frac{4}{3} = \frac{\left(x + 2\right)^{2}}{4} \\ 30 & \displaystyle y - \frac{5}{12} = \left(x - 1\right)^{2} \\ 31 & \displaystyle y + \frac{10}{3} = \frac{\left(x + 1\right)^{2}}{4} \\ 32 & \displaystyle y + \frac{19}{12} = \left(x + 2\right)^{2} \\ 33 & \displaystyle y + \frac{3}{4} = \frac{\left(x + 2\right)^{2}}{3} \\ 34 & \displaystyle y + \frac{13}{3} = \frac{\left(x + 2\right)^{2}}{4} \\ 35 & \displaystyle y - \frac{5}{12} = \left(x + 1\right)^{2} \\ 36 & \displaystyle y + \frac{1}{3} = \frac{\left(x + 1\right)^{2}}{4} \\ 37 & \displaystyle y - \frac{5}{6} = \frac{\left(x - 2\right)^{2}}{2} \\ 38 & \displaystyle y + \frac{13}{6} = \frac{\left(x - 2\right)^{2}}{2} \\ 39 & \displaystyle y + \frac{13}{3} = \frac{\left(x - 2\right)^{2}}{4} \\ 40 & \displaystyle y + \frac{11}{4} = \frac{\left(x - 1\right)^{2}}{3} \\ 41 & \displaystyle y + \frac{10}{3} = \frac{\left(x - 1\right)^{2}}{4} \\ 42 & \displaystyle y - \frac{1}{4} = \frac{\left(x - 1\right)^{2}}{3} \\ \hline \end{array} \]

• Vamos resolver o item 1. Os passos são os mesmos para todos os itens; só mudam os valores.

• A equação é

  \[x^{2} + 2 x - \frac{27 y}{13} - \frac{217}{26} = 0\]

• Vamos separar os termos em \(x\):

  \[x^{2} + 2 x\]

• Completando o quadrado, isto é igual a:

  \[\left(x + 1\right)^{2} - 1\]

• Agora, somamos esta expressão com o termo em \(y\) e o termo independente. O resultado é uma equação equivalente à original:

  \[- \frac{27 y}{13} + \left(x + 1\right)^{2} - \frac{243}{26} = 0\]

• Rearrumando, chegamos à forma canônica:

  \[y + \frac{9}{2} = \frac{13 \left(x + 1\right)^{2}}{27}\]

As respostas para todos os itens são:

\[ \begin{array}{r|l} \hline 1 & \displaystyle y + \frac{9}{2} = \frac{13 \left(x + 1\right)^{2}}{27} \\ 2 & \displaystyle y + \frac{17}{6} = \frac{13 \left(x + 1\right)^{2}}{9} \\ 3 & \displaystyle y + \frac{1}{2} = \frac{13 \left(x - 2\right)^{2}}{27} \\ 4 & \displaystyle y + \frac{11}{6} = \frac{13 \left(x - 1\right)^{2}}{9} \\ 5 & \displaystyle y + \frac{2}{3} = \frac{13 \left(x - 2\right)^{2}}{18} \\ 6 & \displaystyle y + \frac{2}{3} = \frac{13 \left(x - 1\right)^{2}}{18} \\ 7 & \displaystyle y + \frac{9}{2} = \frac{13 \left(x + 2\right)^{2}}{27} \\ 8 & \displaystyle y + \frac{4}{3} = \frac{13 \left(x - 2\right)^{2}}{36} \\ 9 & \displaystyle y - \frac{7}{6} = \frac{13 \left(x + 2\right)^{2}}{9} \\ 10 & \displaystyle y + \frac{1}{2} = \frac{13 \left(x + 1\right)^{2}}{27} \\ 11 & \displaystyle y + \frac{7}{3} = \frac{13 \left(x - 2\right)^{2}}{36} \\ 12 & \displaystyle y + \frac{17}{6} = \frac{13 \left(x - 2\right)^{2}}{9} \\ 13 & \displaystyle y + \frac{2}{3} = \frac{13 \left(x + 2\right)^{2}}{18} \\ 14 & \displaystyle y + \frac{3}{2} = \frac{13 \left(x - 1\right)^{2}}{27} \\ 15 & \displaystyle y + \frac{3}{2} = \frac{13 \left(x + 1\right)^{2}}{27} \\ 16 & \displaystyle y + \frac{11}{3} = \frac{13 \left(x - 1\right)^{2}}{18} \\ 17 & \displaystyle y + \frac{7}{3} = \frac{13 \left(x - 1\right)^{2}}{36} \\ 18 & \displaystyle y + \frac{9}{2} = \frac{13 \left(x - 1\right)^{2}}{27} \\ 19 & \displaystyle y - \frac{7}{6} = \frac{13 \left(x - 1\right)^{2}}{9} \\ 20 & \displaystyle y + \frac{8}{3} = \frac{13 \left(x + 1\right)^{2}}{18} \\ 21 & \displaystyle y + \frac{3}{2} = \frac{13 \left(x - 2\right)^{2}}{27} \\ 22 & \displaystyle y + \frac{7}{2} = \frac{13 \left(x + 1\right)^{2}}{27} \\ 23 & \displaystyle y + \frac{7}{2} = \frac{13 \left(x - 2\right)^{2}}{27} \\ 24 & \displaystyle y + \frac{2}{3} = \frac{13 \left(x + 1\right)^{2}}{18} \\ 25 & \displaystyle y + \frac{7}{2} = \frac{13 \left(x + 2\right)^{2}}{27} \\ 26 & \displaystyle y + \frac{1}{2} = \frac{13 \left(x + 2\right)^{2}}{27} \\ 27 & \displaystyle y - \frac{7}{6} = \frac{13 \left(x - 2\right)^{2}}{9} \\ 28 & \displaystyle y + \frac{9}{2} = \frac{13 \left(x - 2\right)^{2}}{27} \\ 29 & \displaystyle y + \frac{7}{3} = \frac{13 \left(x + 2\right)^{2}}{36} \\ 30 & \displaystyle y - \frac{1}{6} = \frac{13 \left(x - 1\right)^{2}}{9} \\ 31 & \displaystyle y + \frac{13}{3} = \frac{13 \left(x + 1\right)^{2}}{36} \\ 32 & \displaystyle y + \frac{11}{6} = \frac{13 \left(x + 2\right)^{2}}{9} \\ 33 & \displaystyle y + \frac{3}{2} = \frac{13 \left(x + 2\right)^{2}}{27} \\ 34 & \displaystyle y + \frac{16}{3} = \frac{13 \left(x + 2\right)^{2}}{36} \\ 35 & \displaystyle y - \frac{1}{6} = \frac{13 \left(x + 1\right)^{2}}{9} \\ 36 & \displaystyle y + \frac{4}{3} = \frac{13 \left(x + 1\right)^{2}}{36} \\ 37 & \displaystyle y - \frac{1}{3} = \frac{13 \left(x - 2\right)^{2}}{18} \\ 38 & \displaystyle y + \frac{8}{3} = \frac{13 \left(x - 2\right)^{2}}{18} \\ 39 & \displaystyle y + \frac{16}{3} = \frac{13 \left(x - 2\right)^{2}}{36} \\ 40 & \displaystyle y + \frac{7}{2} = \frac{13 \left(x - 1\right)^{2}}{27} \\ 41 & \displaystyle y + \frac{13}{3} = \frac{13 \left(x - 1\right)^{2}}{36} \\ 42 & \displaystyle y + \frac{1}{2} = \frac{13 \left(x - 1\right)^{2}}{27} \\ \hline \end{array} \]

• Vamos resolver o item 1. Os passos são os mesmos para todos os itens; só mudam os valores.

• Os pontos no interior da boca são exatamente aqueles que estão, ao mesmo tempo, abaixo da parte superior e acima da parte inferior.

• São os pontos \((x, y)\) que satisfazem o seguinte sistema de inequações:

  \[ \begin{cases} y + \frac{15}{4} < \frac{\left(x + 1\right)^{2}}{3} \\ y + \frac{9}{2} > \frac{13 \left(x + 1\right)^{2}}{27} \end{cases} \]

As respostas para todos os itens são:

\[ \begin{array}{r|l} \hline 1 & \begin{cases} \displaystyle y + \frac{15}{4} < \frac{\left(x + 1\right)^{2}}{3} \\ \displaystyle y + \frac{9}{2} > \frac{13 \left(x + 1\right)^{2}}{27} \end{cases} \\ 2 & \begin{cases} \displaystyle y + \frac{31}{12} < \left(x + 1\right)^{2} \\ \displaystyle y + \frac{17}{6} > \frac{13 \left(x + 1\right)^{2}}{9} \end{cases} \\ 3 & \begin{cases} \displaystyle y - \frac{1}{4} < \frac{\left(x - 2\right)^{2}}{3} \\ \displaystyle y + \frac{1}{2} > \frac{13 \left(x - 2\right)^{2}}{27} \end{cases} \\ 4 & \begin{cases} \displaystyle y + \frac{19}{12} < \left(x - 1\right)^{2} \\ \displaystyle y + \frac{11}{6} > \frac{13 \left(x - 1\right)^{2}}{9} \end{cases} \\ 5 & \begin{cases} \displaystyle y + \frac{1}{6} < \frac{\left(x - 2\right)^{2}}{2} \\ \displaystyle y + \frac{2}{3} > \frac{13 \left(x - 2\right)^{2}}{18} \end{cases} \\ 6 & \begin{cases} \displaystyle y + \frac{1}{6} < \frac{\left(x - 1\right)^{2}}{2} \\ \displaystyle y + \frac{2}{3} > \frac{13 \left(x - 1\right)^{2}}{18} \end{cases} \\ 7 & \begin{cases} \displaystyle y + \frac{15}{4} < \frac{\left(x + 2\right)^{2}}{3} \\ \displaystyle y + \frac{9}{2} > \frac{13 \left(x + 2\right)^{2}}{27} \end{cases} \\ 8 & \begin{cases} \displaystyle y + \frac{1}{3} < \frac{\left(x - 2\right)^{2}}{4} \\ \displaystyle y + \frac{4}{3} > \frac{13 \left(x - 2\right)^{2}}{36} \end{cases} \\ 9 & \begin{cases} \displaystyle y - \frac{17}{12} < \left(x + 2\right)^{2} \\ \displaystyle y - \frac{7}{6} > \frac{13 \left(x + 2\right)^{2}}{9} \end{cases} \\ 10 & \begin{cases} \displaystyle y - \frac{1}{4} < \frac{\left(x + 1\right)^{2}}{3} \\ \displaystyle y + \frac{1}{2} > \frac{13 \left(x + 1\right)^{2}}{27} \end{cases} \\ 11 & \begin{cases} \displaystyle y + \frac{4}{3} < \frac{\left(x - 2\right)^{2}}{4} \\ \displaystyle y + \frac{7}{3} > \frac{13 \left(x - 2\right)^{2}}{36} \end{cases} \\ 12 & \begin{cases} \displaystyle y + \frac{31}{12} < \left(x - 2\right)^{2} \\ \displaystyle y + \frac{17}{6} > \frac{13 \left(x - 2\right)^{2}}{9} \end{cases} \\ 13 & \begin{cases} \displaystyle y + \frac{1}{6} < \frac{\left(x + 2\right)^{2}}{2} \\ \displaystyle y + \frac{2}{3} > \frac{13 \left(x + 2\right)^{2}}{18} \end{cases} \\ 14 & \begin{cases} \displaystyle y + \frac{3}{4} < \frac{\left(x - 1\right)^{2}}{3} \\ \displaystyle y + \frac{3}{2} > \frac{13 \left(x - 1\right)^{2}}{27} \end{cases} \\ 15 & \begin{cases} \displaystyle y + \frac{3}{4} < \frac{\left(x + 1\right)^{2}}{3} \\ \displaystyle y + \frac{3}{2} > \frac{13 \left(x + 1\right)^{2}}{27} \end{cases} \\ 16 & \begin{cases} \displaystyle y + \frac{19}{6} < \frac{\left(x - 1\right)^{2}}{2} \\ \displaystyle y + \frac{11}{3} > \frac{13 \left(x - 1\right)^{2}}{18} \end{cases} \\ 17 & \begin{cases} \displaystyle y + \frac{4}{3} < \frac{\left(x - 1\right)^{2}}{4} \\ \displaystyle y + \frac{7}{3} > \frac{13 \left(x - 1\right)^{2}}{36} \end{cases} \\ 18 & \begin{cases} \displaystyle y + \frac{15}{4} < \frac{\left(x - 1\right)^{2}}{3} \\ \displaystyle y + \frac{9}{2} > \frac{13 \left(x - 1\right)^{2}}{27} \end{cases} \\ 19 & \begin{cases} \displaystyle y - \frac{17}{12} < \left(x - 1\right)^{2} \\ \displaystyle y - \frac{7}{6} > \frac{13 \left(x - 1\right)^{2}}{9} \end{cases} \\ 20 & \begin{cases} \displaystyle y + \frac{13}{6} < \frac{\left(x + 1\right)^{2}}{2} \\ \displaystyle y + \frac{8}{3} > \frac{13 \left(x + 1\right)^{2}}{18} \end{cases} \\ 21 & \begin{cases} \displaystyle y + \frac{3}{4} < \frac{\left(x - 2\right)^{2}}{3} \\ \displaystyle y + \frac{3}{2} > \frac{13 \left(x - 2\right)^{2}}{27} \end{cases} \\ 22 & \begin{cases} \displaystyle y + \frac{11}{4} < \frac{\left(x + 1\right)^{2}}{3} \\ \displaystyle y + \frac{7}{2} > \frac{13 \left(x + 1\right)^{2}}{27} \end{cases} \\ 23 & \begin{cases} \displaystyle y + \frac{11}{4} < \frac{\left(x - 2\right)^{2}}{3} \\ \displaystyle y + \frac{7}{2} > \frac{13 \left(x - 2\right)^{2}}{27} \end{cases} \\ 24 & \begin{cases} \displaystyle y + \frac{1}{6} < \frac{\left(x + 1\right)^{2}}{2} \\ \displaystyle y + \frac{2}{3} > \frac{13 \left(x + 1\right)^{2}}{18} \end{cases} \\ 25 & \begin{cases} \displaystyle y + \frac{11}{4} < \frac{\left(x + 2\right)^{2}}{3} \\ \displaystyle y + \frac{7}{2} > \frac{13 \left(x + 2\right)^{2}}{27} \end{cases} \\ 26 & \begin{cases} \displaystyle y - \frac{1}{4} < \frac{\left(x + 2\right)^{2}}{3} \\ \displaystyle y + \frac{1}{2} > \frac{13 \left(x + 2\right)^{2}}{27} \end{cases} \\ 27 & \begin{cases} \displaystyle y - \frac{17}{12} < \left(x - 2\right)^{2} \\ \displaystyle y - \frac{7}{6} > \frac{13 \left(x - 2\right)^{2}}{9} \end{cases} \\ 28 & \begin{cases} \displaystyle y + \frac{15}{4} < \frac{\left(x - 2\right)^{2}}{3} \\ \displaystyle y + \frac{9}{2} > \frac{13 \left(x - 2\right)^{2}}{27} \end{cases} \\ 29 & \begin{cases} \displaystyle y + \frac{4}{3} < \frac{\left(x + 2\right)^{2}}{4} \\ \displaystyle y + \frac{7}{3} > \frac{13 \left(x + 2\right)^{2}}{36} \end{cases} \\ 30 & \begin{cases} \displaystyle y - \frac{5}{12} < \left(x - 1\right)^{2} \\ \displaystyle y - \frac{1}{6} > \frac{13 \left(x - 1\right)^{2}}{9} \end{cases} \\ 31 & \begin{cases} \displaystyle y + \frac{10}{3} < \frac{\left(x + 1\right)^{2}}{4} \\ \displaystyle y + \frac{13}{3} > \frac{13 \left(x + 1\right)^{2}}{36} \end{cases} \\ 32 & \begin{cases} \displaystyle y + \frac{19}{12} < \left(x + 2\right)^{2} \\ \displaystyle y + \frac{11}{6} > \frac{13 \left(x + 2\right)^{2}}{9} \end{cases} \\ 33 & \begin{cases} \displaystyle y + \frac{3}{4} < \frac{\left(x + 2\right)^{2}}{3} \\ \displaystyle y + \frac{3}{2} > \frac{13 \left(x + 2\right)^{2}}{27} \end{cases} \\ 34 & \begin{cases} \displaystyle y + \frac{13}{3} < \frac{\left(x + 2\right)^{2}}{4} \\ \displaystyle y + \frac{16}{3} > \frac{13 \left(x + 2\right)^{2}}{36} \end{cases} \\ 35 & \begin{cases} \displaystyle y - \frac{5}{12} < \left(x + 1\right)^{2} \\ \displaystyle y - \frac{1}{6} > \frac{13 \left(x + 1\right)^{2}}{9} \end{cases} \\ 36 & \begin{cases} \displaystyle y + \frac{1}{3} < \frac{\left(x + 1\right)^{2}}{4} \\ \displaystyle y + \frac{4}{3} > \frac{13 \left(x + 1\right)^{2}}{36} \end{cases} \\ 37 & \begin{cases} \displaystyle y - \frac{5}{6} < \frac{\left(x - 2\right)^{2}}{2} \\ \displaystyle y - \frac{1}{3} > \frac{13 \left(x - 2\right)^{2}}{18} \end{cases} \\ 38 & \begin{cases} \displaystyle y + \frac{13}{6} < \frac{\left(x - 2\right)^{2}}{2} \\ \displaystyle y + \frac{8}{3} > \frac{13 \left(x - 2\right)^{2}}{18} \end{cases} \\ 39 & \begin{cases} \displaystyle y + \frac{13}{3} < \frac{\left(x - 2\right)^{2}}{4} \\ \displaystyle y + \frac{16}{3} > \frac{13 \left(x - 2\right)^{2}}{36} \end{cases} \\ 40 & \begin{cases} \displaystyle y + \frac{11}{4} < \frac{\left(x - 1\right)^{2}}{3} \\ \displaystyle y + \frac{7}{2} > \frac{13 \left(x - 1\right)^{2}}{27} \end{cases} \\ 41 & \begin{cases} \displaystyle y + \frac{10}{3} < \frac{\left(x - 1\right)^{2}}{4} \\ \displaystyle y + \frac{13}{3} > \frac{13 \left(x - 1\right)^{2}}{36} \end{cases} \\ 42 & \begin{cases} \displaystyle y - \frac{1}{4} < \frac{\left(x - 1\right)^{2}}{3} \\ \displaystyle y + \frac{1}{2} > \frac{13 \left(x - 1\right)^{2}}{27} \end{cases} \\ \hline \end{array} \]

• Vamos resolver o item 1. Os passos são os mesmos para todos os itens; só mudam os valores.

• A equação canônica de uma hipérbole de eixo real horizontal tem a forma

  \[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]

• Aqui, o ponto \((h, k)\) é o centro, que não foi dado.

• Mas, se você prestar atenção, vai notar que os vértices da hipérbole estão na mesma horizontal \((y = -2)\) que o centro do rosto.

• Então, como as orelhas são simétricas em relação ao centro do rosto, o centro da hipérbole é igual ao centro do círculo do rosto: \((h, k) = (-1, -2)\).

• A excentricidade de uma hipérbole é \(e = c/a\). O enunciado diz que \(e = 11/10\).

• Daí, \(c/a = 11/10\), o que equivale a dizer que \(c = 11a/10\).

• Na hipérbole, \(c^2 = a^2 + b^2\), o que equivale a dizer que \(b^2 = c^2 - a^2\).

• Isto equivale a dizer que \(b^2 = \frac{121a^2}{100} - a^2 = \frac{21a^2}{100}\).

• Levando em conta as informações acima, a hipérbole que procuramos é

  \[ \frac{(x + 1)^2}{a^2} - \frac{(y + 2)^2}{21a^2 / 100} = 1 \]

• Para descobrir o valor de \(a\), podemos usar o fato de que os vértices da hipérbole também pertencem ao círculo do rosto.

• Ou seja, os pontos \((-4, -2)\) e \((2, -2)\) são os vértices da hipérbole.

• Usando o segundo vértice, poderíamos substituir \(x\) por \(2\) e \(y\) por \(-2\) e resolver a equação para achar o valor de \(a\).

• Mas nem precisamos fazer isto. O valor de \(a\) é justamente a distância entre o centro e o vértice da hipérbole, e acabamos de observar que esta distância é o raio do círculo do rosto: ou seja, \(a = 3\).

• A equação da hipérbole, fica, então

  \[ \frac{(x + 1)^2}{9} - \frac{(y + 2)^2}{189 / 100} = 1 \]

As respostas para todos os itens são:

\[ \begin{array}{r|l} \hline 1 & \displaystyle \frac{\left(x + 1\right)^{2}}{9} - \frac{\left(y + 2\right)^{2}}{\frac{189}{100}} = 1 \\ 2 & \displaystyle \frac{\left(x + 1\right)^{2}}{1} - \frac{\left(y + 2\right)^{2}}{\frac{21}{100}} = 1 \\ 3 & \displaystyle \frac{\left(x - 2\right)^{2}}{9} - \frac{\left(y - 2\right)^{2}}{\frac{189}{100}} = 1 \\ 4 & \displaystyle \frac{\left(x - 1\right)^{2}}{1} - \frac{\left(y + 1\right)^{2}}{\frac{21}{100}} = 1 \\ 5 & \displaystyle \frac{\left(x - 2\right)^{2}}{4} - \frac{\left(y - 1\right)^{2}}{\frac{21}{25}} = 1 \\ 6 & \displaystyle \frac{\left(x - 1\right)^{2}}{4} - \frac{\left(y - 1\right)^{2}}{\frac{21}{25}} = 1 \\ 7 & \displaystyle \frac{\left(x + 2\right)^{2}}{9} - \frac{\left(y + 2\right)^{2}}{\frac{189}{100}} = 1 \\ 8 & \displaystyle \frac{\left(x - 2\right)^{2}}{16} - \frac{\left(y - 2\right)^{2}}{\frac{84}{25}} = 1 \\ 9 & \displaystyle \frac{\left(x + 2\right)^{2}}{1} - \frac{\left(y - 2\right)^{2}}{\frac{21}{100}} = 1 \\ 10 & \displaystyle \frac{\left(x + 1\right)^{2}}{9} - \frac{\left(y - 2\right)^{2}}{\frac{189}{100}} = 1 \\ 11 & \displaystyle \frac{\left(x - 2\right)^{2}}{16} - \frac{\left(y - 1\right)^{2}}{\frac{84}{25}} = 1 \\ 12 & \displaystyle \frac{\left(x - 2\right)^{2}}{1} - \frac{\left(y + 2\right)^{2}}{\frac{21}{100}} = 1 \\ 13 & \displaystyle \frac{\left(x + 2\right)^{2}}{4} - \frac{\left(y - 1\right)^{2}}{\frac{21}{25}} = 1 \\ 14 & \displaystyle \frac{\left(x - 1\right)^{2}}{9} - \frac{\left(y - 1\right)^{2}}{\frac{189}{100}} = 1 \\ 15 & \displaystyle \frac{\left(x + 1\right)^{2}}{9} - \frac{\left(y - 1\right)^{2}}{\frac{189}{100}} = 1 \\ 16 & \displaystyle \frac{\left(x - 1\right)^{2}}{4} - \frac{\left(y + 2\right)^{2}}{\frac{21}{25}} = 1 \\ 17 & \displaystyle \frac{\left(x - 1\right)^{2}}{16} - \frac{\left(y - 1\right)^{2}}{\frac{84}{25}} = 1 \\ 18 & \displaystyle \frac{\left(x - 1\right)^{2}}{9} - \frac{\left(y + 2\right)^{2}}{\frac{189}{100}} = 1 \\ 19 & \displaystyle \frac{\left(x - 1\right)^{2}}{1} - \frac{\left(y - 2\right)^{2}}{\frac{21}{100}} = 1 \\ 20 & \displaystyle \frac{\left(x + 1\right)^{2}}{4} - \frac{\left(y + 1\right)^{2}}{\frac{21}{25}} = 1 \\ 21 & \displaystyle \frac{\left(x - 2\right)^{2}}{9} - \frac{\left(y - 1\right)^{2}}{\frac{189}{100}} = 1 \\ 22 & \displaystyle \frac{\left(x + 1\right)^{2}}{9} - \frac{\left(y + 1\right)^{2}}{\frac{189}{100}} = 1 \\ 23 & \displaystyle \frac{\left(x - 2\right)^{2}}{9} - \frac{\left(y + 1\right)^{2}}{\frac{189}{100}} = 1 \\ 24 & \displaystyle \frac{\left(x + 1\right)^{2}}{4} - \frac{\left(y - 1\right)^{2}}{\frac{21}{25}} = 1 \\ 25 & \displaystyle \frac{\left(x + 2\right)^{2}}{9} - \frac{\left(y + 1\right)^{2}}{\frac{189}{100}} = 1 \\ 26 & \displaystyle \frac{\left(x + 2\right)^{2}}{9} - \frac{\left(y - 2\right)^{2}}{\frac{189}{100}} = 1 \\ 27 & \displaystyle \frac{\left(x - 2\right)^{2}}{1} - \frac{\left(y - 2\right)^{2}}{\frac{21}{100}} = 1 \\ 28 & \displaystyle \frac{\left(x - 2\right)^{2}}{9} - \frac{\left(y + 2\right)^{2}}{\frac{189}{100}} = 1 \\ 29 & \displaystyle \frac{\left(x + 2\right)^{2}}{16} - \frac{\left(y - 1\right)^{2}}{\frac{84}{25}} = 1 \\ 30 & \displaystyle \frac{\left(x - 1\right)^{2}}{1} - \frac{\left(y - 1\right)^{2}}{\frac{21}{100}} = 1 \\ 31 & \displaystyle \frac{\left(x + 1\right)^{2}}{16} - \frac{\left(y + 1\right)^{2}}{\frac{84}{25}} = 1 \\ 32 & \displaystyle \frac{\left(x + 2\right)^{2}}{1} - \frac{\left(y + 1\right)^{2}}{\frac{21}{100}} = 1 \\ 33 & \displaystyle \frac{\left(x + 2\right)^{2}}{9} - \frac{\left(y - 1\right)^{2}}{\frac{189}{100}} = 1 \\ 34 & \displaystyle \frac{\left(x + 2\right)^{2}}{16} - \frac{\left(y + 2\right)^{2}}{\frac{84}{25}} = 1 \\ 35 & \displaystyle \frac{\left(x + 1\right)^{2}}{1} - \frac{\left(y - 1\right)^{2}}{\frac{21}{100}} = 1 \\ 36 & \displaystyle \frac{\left(x + 1\right)^{2}}{16} - \frac{\left(y - 2\right)^{2}}{\frac{84}{25}} = 1 \\ 37 & \displaystyle \frac{\left(x - 2\right)^{2}}{4} - \frac{\left(y - 2\right)^{2}}{\frac{21}{25}} = 1 \\ 38 & \displaystyle \frac{\left(x - 2\right)^{2}}{4} - \frac{\left(y + 1\right)^{2}}{\frac{21}{25}} = 1 \\ 39 & \displaystyle \frac{\left(x - 2\right)^{2}}{16} - \frac{\left(y + 2\right)^{2}}{\frac{84}{25}} = 1 \\ 40 & \displaystyle \frac{\left(x - 1\right)^{2}}{9} - \frac{\left(y + 1\right)^{2}}{\frac{189}{100}} = 1 \\ 41 & \displaystyle \frac{\left(x - 1\right)^{2}}{16} - \frac{\left(y + 1\right)^{2}}{\frac{84}{25}} = 1 \\ 42 & \displaystyle \frac{\left(x - 1\right)^{2}}{9} - \frac{\left(y - 2\right)^{2}}{\frac{189}{100}} = 1 \\ \hline \end{array} \]

• Vamos resolver o item 1. Os passos são os mesmos para todos os itens; só mudam os valores.

• Lembrando que o centro do círculo tem \(x = -1\), os limites das orelhas são as duas retas verticais

  \[x = -1 - 27/8 = -35/8\]

  e

  \[x = -1 + 27/8 = 19/8\]

• O capítulo \(10\) do livro diz que os pontos que estão nas mesmas regiões que os focos da hipérbole (ou seja, nas orelhas) satisfazem a inequação

  \[ \frac{(x + 1)^2}{9} - \frac{(y + 2)^2}{189 / 100} > 1 \]

• Assim, a parte sombreada das orelhas corresponde aos pontos que satisfazem o sistema de inequações

  \[ \begin{cases} \displaystyle \frac{(x + 1)^2}{9} - \frac{(y + 2)^2}{189 / 100} > 1 \\ \displaystyle x > \frac{-35}{8} \\ \displaystyle x < \frac{19}{8} \end{cases} \]

As respostas para todos os itens são:

\[ \begin{array}{r|l} \hline 1 & \begin{cases} \displaystyle \frac{\left(x + 1\right)^{2}}{9} - \frac{\left(y + 2\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{35}{8} \\ \displaystyle x < \frac{19}{8} \end{cases} \\ 2 & \begin{cases} \displaystyle \frac{\left(x + 1\right)^{2}}{1} - \frac{\left(y + 2\right)^{2}}{\frac{21}{100}} > 1 \\ \displaystyle x > - \frac{17}{8} \\ \displaystyle x < \frac{1}{8} \end{cases} \\ 3 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{9} - \frac{\left(y - 2\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{11}{8} \\ \displaystyle x < \frac{43}{8} \end{cases} \\ 4 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{1} - \frac{\left(y + 1\right)^{2}}{\frac{21}{100}} > 1 \\ \displaystyle x > - \frac{1}{8} \\ \displaystyle x < \frac{17}{8} \end{cases} \\ 5 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{4} - \frac{\left(y - 1\right)^{2}}{\frac{21}{25}} > 1 \\ \displaystyle x > - \frac{1}{4} \\ \displaystyle x < \frac{17}{4} \end{cases} \\ 6 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{4} - \frac{\left(y - 1\right)^{2}}{\frac{21}{25}} > 1 \\ \displaystyle x > - \frac{5}{4} \\ \displaystyle x < \frac{13}{4} \end{cases} \\ 7 & \begin{cases} \displaystyle \frac{\left(x + 2\right)^{2}}{9} - \frac{\left(y + 2\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{43}{8} \\ \displaystyle x < \frac{11}{8} \end{cases} \\ 8 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{16} - \frac{\left(y - 2\right)^{2}}{\frac{84}{25}} > 1 \\ \displaystyle x > - \frac{5}{2} \\ \displaystyle x < \frac{13}{2} \end{cases} \\ 9 & \begin{cases} \displaystyle \frac{\left(x + 2\right)^{2}}{1} - \frac{\left(y - 2\right)^{2}}{\frac{21}{100}} > 1 \\ \displaystyle x > - \frac{25}{8} \\ \displaystyle x < - \frac{7}{8} \end{cases} \\ 10 & \begin{cases} \displaystyle \frac{\left(x + 1\right)^{2}}{9} - \frac{\left(y - 2\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{35}{8} \\ \displaystyle x < \frac{19}{8} \end{cases} \\ 11 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{16} - \frac{\left(y - 1\right)^{2}}{\frac{84}{25}} > 1 \\ \displaystyle x > - \frac{5}{2} \\ \displaystyle x < \frac{13}{2} \end{cases} \\ 12 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{1} - \frac{\left(y + 2\right)^{2}}{\frac{21}{100}} > 1 \\ \displaystyle x > \frac{7}{8} \\ \displaystyle x < \frac{25}{8} \end{cases} \\ 13 & \begin{cases} \displaystyle \frac{\left(x + 2\right)^{2}}{4} - \frac{\left(y - 1\right)^{2}}{\frac{21}{25}} > 1 \\ \displaystyle x > - \frac{17}{4} \\ \displaystyle x < \frac{1}{4} \end{cases} \\ 14 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{9} - \frac{\left(y - 1\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{19}{8} \\ \displaystyle x < \frac{35}{8} \end{cases} \\ 15 & \begin{cases} \displaystyle \frac{\left(x + 1\right)^{2}}{9} - \frac{\left(y - 1\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{35}{8} \\ \displaystyle x < \frac{19}{8} \end{cases} \\ 16 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{4} - \frac{\left(y + 2\right)^{2}}{\frac{21}{25}} > 1 \\ \displaystyle x > - \frac{5}{4} \\ \displaystyle x < \frac{13}{4} \end{cases} \\ 17 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{16} - \frac{\left(y - 1\right)^{2}}{\frac{84}{25}} > 1 \\ \displaystyle x > - \frac{7}{2} \\ \displaystyle x < \frac{11}{2} \end{cases} \\ 18 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{9} - \frac{\left(y + 2\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{19}{8} \\ \displaystyle x < \frac{35}{8} \end{cases} \\ 19 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{1} - \frac{\left(y - 2\right)^{2}}{\frac{21}{100}} > 1 \\ \displaystyle x > - \frac{1}{8} \\ \displaystyle x < \frac{17}{8} \end{cases} \\ 20 & \begin{cases} \displaystyle \frac{\left(x + 1\right)^{2}}{4} - \frac{\left(y + 1\right)^{2}}{\frac{21}{25}} > 1 \\ \displaystyle x > - \frac{13}{4} \\ \displaystyle x < \frac{5}{4} \end{cases} \\ 21 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{9} - \frac{\left(y - 1\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{11}{8} \\ \displaystyle x < \frac{43}{8} \end{cases} \\ 22 & \begin{cases} \displaystyle \frac{\left(x + 1\right)^{2}}{9} - \frac{\left(y + 1\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{35}{8} \\ \displaystyle x < \frac{19}{8} \end{cases} \\ 23 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{9} - \frac{\left(y + 1\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{11}{8} \\ \displaystyle x < \frac{43}{8} \end{cases} \\ 24 & \begin{cases} \displaystyle \frac{\left(x + 1\right)^{2}}{4} - \frac{\left(y - 1\right)^{2}}{\frac{21}{25}} > 1 \\ \displaystyle x > - \frac{13}{4} \\ \displaystyle x < \frac{5}{4} \end{cases} \\ 25 & \begin{cases} \displaystyle \frac{\left(x + 2\right)^{2}}{9} - \frac{\left(y + 1\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{43}{8} \\ \displaystyle x < \frac{11}{8} \end{cases} \\ 26 & \begin{cases} \displaystyle \frac{\left(x + 2\right)^{2}}{9} - \frac{\left(y - 2\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{43}{8} \\ \displaystyle x < \frac{11}{8} \end{cases} \\ 27 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{1} - \frac{\left(y - 2\right)^{2}}{\frac{21}{100}} > 1 \\ \displaystyle x > \frac{7}{8} \\ \displaystyle x < \frac{25}{8} \end{cases} \\ 28 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{9} - \frac{\left(y + 2\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{11}{8} \\ \displaystyle x < \frac{43}{8} \end{cases} \\ 29 & \begin{cases} \displaystyle \frac{\left(x + 2\right)^{2}}{16} - \frac{\left(y - 1\right)^{2}}{\frac{84}{25}} > 1 \\ \displaystyle x > - \frac{13}{2} \\ \displaystyle x < \frac{5}{2} \end{cases} \\ 30 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{1} - \frac{\left(y - 1\right)^{2}}{\frac{21}{100}} > 1 \\ \displaystyle x > - \frac{1}{8} \\ \displaystyle x < \frac{17}{8} \end{cases} \\ 31 & \begin{cases} \displaystyle \frac{\left(x + 1\right)^{2}}{16} - \frac{\left(y + 1\right)^{2}}{\frac{84}{25}} > 1 \\ \displaystyle x > - \frac{11}{2} \\ \displaystyle x < \frac{7}{2} \end{cases} \\ 32 & \begin{cases} \displaystyle \frac{\left(x + 2\right)^{2}}{1} - \frac{\left(y + 1\right)^{2}}{\frac{21}{100}} > 1 \\ \displaystyle x > - \frac{25}{8} \\ \displaystyle x < - \frac{7}{8} \end{cases} \\ 33 & \begin{cases} \displaystyle \frac{\left(x + 2\right)^{2}}{9} - \frac{\left(y - 1\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{43}{8} \\ \displaystyle x < \frac{11}{8} \end{cases} \\ 34 & \begin{cases} \displaystyle \frac{\left(x + 2\right)^{2}}{16} - \frac{\left(y + 2\right)^{2}}{\frac{84}{25}} > 1 \\ \displaystyle x > - \frac{13}{2} \\ \displaystyle x < \frac{5}{2} \end{cases} \\ 35 & \begin{cases} \displaystyle \frac{\left(x + 1\right)^{2}}{1} - \frac{\left(y - 1\right)^{2}}{\frac{21}{100}} > 1 \\ \displaystyle x > - \frac{17}{8} \\ \displaystyle x < \frac{1}{8} \end{cases} \\ 36 & \begin{cases} \displaystyle \frac{\left(x + 1\right)^{2}}{16} - \frac{\left(y - 2\right)^{2}}{\frac{84}{25}} > 1 \\ \displaystyle x > - \frac{11}{2} \\ \displaystyle x < \frac{7}{2} \end{cases} \\ 37 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{4} - \frac{\left(y - 2\right)^{2}}{\frac{21}{25}} > 1 \\ \displaystyle x > - \frac{1}{4} \\ \displaystyle x < \frac{17}{4} \end{cases} \\ 38 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{4} - \frac{\left(y + 1\right)^{2}}{\frac{21}{25}} > 1 \\ \displaystyle x > - \frac{1}{4} \\ \displaystyle x < \frac{17}{4} \end{cases} \\ 39 & \begin{cases} \displaystyle \frac{\left(x - 2\right)^{2}}{16} - \frac{\left(y + 2\right)^{2}}{\frac{84}{25}} > 1 \\ \displaystyle x > - \frac{5}{2} \\ \displaystyle x < \frac{13}{2} \end{cases} \\ 40 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{9} - \frac{\left(y + 1\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{19}{8} \\ \displaystyle x < \frac{35}{8} \end{cases} \\ 41 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{16} - \frac{\left(y + 1\right)^{2}}{\frac{84}{25}} > 1 \\ \displaystyle x > - \frac{7}{2} \\ \displaystyle x < \frac{11}{2} \end{cases} \\ 42 & \begin{cases} \displaystyle \frac{\left(x - 1\right)^{2}}{9} - \frac{\left(y - 2\right)^{2}}{\frac{189}{100}} > 1 \\ \displaystyle x > - \frac{19}{8} \\ \displaystyle x < \frac{35}{8} \end{cases} \\ \hline \end{array} \]