Fique à vontade para consultar os coleguinhas e para usar programas como o Geogebra, mas somente soluções analíticas serão aceitas — nada de responder no olhômetro.
Uma solução analítica é aquela em que você detalha todos os passos intermediários: não vale resolver tudo no Geogebra e apresentar o resultado final; é preciso mostrar o passo-a-passo.
Entregue (via Moodle) sua resolução escrita no formato que você preferir: manuscrito escaneado ou fotografado, documento gerado via \(\LaTeX\) etc. O importante é que a resolução esteja legível. Se você for fotografar sua resolução, use um aplicativo como Clear Scan para gerar um resultado melhor.
Além da resolução por escrito, entregue também (via Moodle) um arquivo contendo um vídeo de no máximo 5 minutos onde você explica em detalhes a resolução de uma parte da sua questão.
Bom trabalho.
Os dados da sua questão dependem do valor de \(n\) sorteado para você.
Veja o seu valor de \(n\) nesta lista.
Você vai achar equações de cônicas que são o rosto, os olhos, o nariz, a boca e as orelhas de um emoji no \(\mathbb{R}^2\).
Além disso, você vai achar inequações envolvendo cônicas que correspondem às áreas preenchidas da boca e das orelhas.
Em todos os seus cálculos e respostas, use frações e radicais.
Não use valores numéricos com vírgulas decimais em momento algum.
Seu emoji vai ficar assim:
O rosto é o círculo de equação geral dada abaixo (veja o seu número \(n\) nesta lista).
Ache a equação canônica deste círculo.
\(\displaystyle \quad x^{2} + 2 x + y^{2} + 4 y - 4 = 0\)
\(\displaystyle \quad x^{2} + 2 x + y^{2} + 4 y + 4 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} - 4 y - 1 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} + 2 y + 1 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} - 2 y + 1 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} - 2 y - 2 = 0\)
\(\displaystyle \quad x^{2} + 4 x + y^{2} + 4 y - 1 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} - 4 y - 8 = 0\)
\(\displaystyle \quad x^{2} + 4 x + y^{2} - 4 y + 7 = 0\)
\(\displaystyle \quad x^{2} + 2 x + y^{2} - 4 y - 4 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} - 2 y - 11 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} + 4 y + 7 = 0\)
\(\displaystyle \quad x^{2} + 4 x + y^{2} - 2 y + 1 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} - 2 y - 7 = 0\)
\(\displaystyle \quad x^{2} + 2 x + y^{2} - 2 y - 7 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} + 4 y + 1 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} - 2 y - 14 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} + 4 y - 4 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} - 4 y + 4 = 0\)
\(\displaystyle \quad x^{2} + 2 x + y^{2} + 2 y - 2 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} - 2 y - 4 = 0\)
\(\displaystyle \quad x^{2} + 2 x + y^{2} + 2 y - 7 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} + 2 y - 4 = 0\)
\(\displaystyle \quad x^{2} + 2 x + y^{2} - 2 y - 2 = 0\)
\(\displaystyle \quad x^{2} + 4 x + y^{2} + 2 y - 4 = 0\)
\(\displaystyle \quad x^{2} + 4 x + y^{2} - 4 y - 1 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} - 4 y + 7 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} + 4 y - 1 = 0\)
\(\displaystyle \quad x^{2} + 4 x + y^{2} - 2 y - 11 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} - 2 y + 1 = 0\)
\(\displaystyle \quad x^{2} + 2 x + y^{2} + 2 y - 14 = 0\)
\(\displaystyle \quad x^{2} + 4 x + y^{2} + 2 y + 4 = 0\)
\(\displaystyle \quad x^{2} + 4 x + y^{2} - 2 y - 4 = 0\)
\(\displaystyle \quad x^{2} + 4 x + y^{2} + 4 y - 8 = 0\)
\(\displaystyle \quad x^{2} + 2 x + y^{2} - 2 y + 1 = 0\)
\(\displaystyle \quad x^{2} + 2 x + y^{2} - 4 y - 11 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} - 4 y + 4 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} + 2 y + 1 = 0\)
\(\displaystyle \quad x^{2} - 4 x + y^{2} + 4 y - 8 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} + 2 y - 7 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} + 2 y - 14 = 0\)
\(\displaystyle \quad x^{2} - 2 x + y^{2} - 4 y - 4 = 0\)
O nariz é a elipse de eixo maior horizontal com as coordenadas do centro, valores de \(a\) e de \(c\) dados abaixo (veja o seu número \(n\) nesta lista).
Ache a equação canônica desta elipse.
\(\displaystyle \quad \text{Centro} = \left( -1; \ -2\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -1; \ -2\right)\ ,\quad a = \frac{11}{100}\ ,\quad c = \frac{1}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ 2\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ -1\right)\ ,\quad a = \frac{11}{100}\ ,\quad c = \frac{1}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ 1\right)\ ,\quad a = \frac{11}{50}\ ,\quad c = \frac{1}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ 1\right)\ ,\quad a = \frac{11}{50}\ ,\quad c = \frac{1}{5}\)
\(\displaystyle \quad \text{Centro} = \left( -2; \ -2\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ 2\right)\ ,\quad a = \frac{11}{25}\ ,\quad c = \frac{2}{5}\)
\(\displaystyle \quad \text{Centro} = \left( -2; \ 2\right)\ ,\quad a = \frac{11}{100}\ ,\quad c = \frac{1}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -1; \ 2\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ 1\right)\ ,\quad a = \frac{11}{25}\ ,\quad c = \frac{2}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ -2\right)\ ,\quad a = \frac{11}{100}\ ,\quad c = \frac{1}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -2; \ 1\right)\ ,\quad a = \frac{11}{50}\ ,\quad c = \frac{1}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ 1\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -1; \ 1\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ -2\right)\ ,\quad a = \frac{11}{50}\ ,\quad c = \frac{1}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ 1\right)\ ,\quad a = \frac{11}{25}\ ,\quad c = \frac{2}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ -2\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ 2\right)\ ,\quad a = \frac{11}{100}\ ,\quad c = \frac{1}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -1; \ -1\right)\ ,\quad a = \frac{11}{50}\ ,\quad c = \frac{1}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ 1\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -1; \ -1\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ -1\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -1; \ 1\right)\ ,\quad a = \frac{11}{50}\ ,\quad c = \frac{1}{5}\)
\(\displaystyle \quad \text{Centro} = \left( -2; \ -1\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -2; \ 2\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ 2\right)\ ,\quad a = \frac{11}{100}\ ,\quad c = \frac{1}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ -2\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -2; \ 1\right)\ ,\quad a = \frac{11}{25}\ ,\quad c = \frac{2}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ 1\right)\ ,\quad a = \frac{11}{100}\ ,\quad c = \frac{1}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -1; \ -1\right)\ ,\quad a = \frac{11}{25}\ ,\quad c = \frac{2}{5}\)
\(\displaystyle \quad \text{Centro} = \left( -2; \ -1\right)\ ,\quad a = \frac{11}{100}\ ,\quad c = \frac{1}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -2; \ 1\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -2; \ -2\right)\ ,\quad a = \frac{11}{25}\ ,\quad c = \frac{2}{5}\)
\(\displaystyle \quad \text{Centro} = \left( -1; \ 1\right)\ ,\quad a = \frac{11}{100}\ ,\quad c = \frac{1}{10}\)
\(\displaystyle \quad \text{Centro} = \left( -1; \ 2\right)\ ,\quad a = \frac{11}{25}\ ,\quad c = \frac{2}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ 2\right)\ ,\quad a = \frac{11}{50}\ ,\quad c = \frac{1}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ -1\right)\ ,\quad a = \frac{11}{50}\ ,\quad c = \frac{1}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 2; \ -2\right)\ ,\quad a = \frac{11}{25}\ ,\quad c = \frac{2}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ -1\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ -1\right)\ ,\quad a = \frac{11}{25}\ ,\quad c = \frac{2}{5}\)
\(\displaystyle \quad \text{Centro} = \left( 1; \ 2\right)\ ,\quad a = \frac{33}{100}\ ,\quad c = \frac{3}{10}\)
Os centros dos olhos são os focos da elipse com equação geral dada abaixo (veja o seu número \(n\) nesta lista).
Ache a equação canônica e as coordenadas dos focos desta elipse.
\(\displaystyle \quad x^{2} + 2 x + \frac{121 y^{2}}{21} + \frac{242 y}{15} + \frac{2297}{240} = 0\)
\(\displaystyle \quad x^{2} + 2 x + \frac{121 y^{2}}{21} + \frac{726 y}{35} + \frac{2169}{112} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} - \frac{3146 y}{105} + \frac{67583}{1680} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} + \frac{968 y}{105} + \frac{7367}{1680} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} - \frac{242 y}{15} + \frac{169}{12} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} - \frac{242 y}{15} + \frac{133}{12} = 0\)
\(\displaystyle \quad x^{2} + 4 x + \frac{121 y^{2}}{21} + \frac{242 y}{15} + \frac{3017}{240} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} - \frac{484 y}{15} + \frac{133}{3} = 0\)
\(\displaystyle \quad x^{2} + 4 x + \frac{121 y^{2}}{21} - \frac{2662 y}{105} + \frac{53063}{1680} = 0\)
\(\displaystyle \quad x^{2} + 2 x + \frac{121 y^{2}}{21} - \frac{3146 y}{105} + \frac{62543}{1680} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} - \frac{726 y}{35} + \frac{624}{35} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} + \frac{726 y}{35} + \frac{2505}{112} = 0\)
\(\displaystyle \quad x^{2} + 4 x + \frac{121 y^{2}}{21} - \frac{242 y}{15} + \frac{169}{12} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} - \frac{1936 y}{105} + \frac{21887}{1680} = 0\)
\(\displaystyle \quad x^{2} + 2 x + \frac{121 y^{2}}{21} - \frac{1936 y}{105} + \frac{21887}{1680} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} + \frac{1936 y}{105} + \frac{6107}{420} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} - \frac{726 y}{35} + \frac{519}{35} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} + \frac{242 y}{15} + \frac{2297}{240} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} - \frac{2662 y}{105} + \frac{48023}{1680} = 0\)
\(\displaystyle \quad x^{2} + 2 x + \frac{121 y^{2}}{21} + \frac{242 y}{35} + \frac{261}{140} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} - \frac{1936 y}{105} + \frac{26927}{1680} = 0\)
\(\displaystyle \quad x^{2} + 2 x + \frac{121 y^{2}}{21} + \frac{484 y}{105} - \frac{269}{336} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} + \frac{484 y}{105} + \frac{739}{336} = 0\)
\(\displaystyle \quad x^{2} + 2 x + \frac{121 y^{2}}{21} - \frac{242 y}{15} + \frac{133}{12} = 0\)
\(\displaystyle \quad x^{2} + 4 x + \frac{121 y^{2}}{21} + \frac{484 y}{105} + \frac{739}{336} = 0\)
\(\displaystyle \quad x^{2} + 4 x + \frac{121 y^{2}}{21} - \frac{3146 y}{105} + \frac{67583}{1680} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} - \frac{2662 y}{105} + \frac{53063}{1680} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} + \frac{242 y}{15} + \frac{3017}{240} = 0\)
\(\displaystyle \quad x^{2} + 4 x + \frac{121 y^{2}}{21} - \frac{726 y}{35} + \frac{624}{35} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} - \frac{484 y}{35} + \frac{5037}{560} = 0\)
\(\displaystyle \quad x^{2} + 2 x + \frac{121 y^{2}}{21} + \frac{242 y}{105} - \frac{379}{105} = 0\)
\(\displaystyle \quad x^{2} + 4 x + \frac{121 y^{2}}{21} + \frac{968 y}{105} + \frac{12407}{1680} = 0\)
\(\displaystyle \quad x^{2} + 4 x + \frac{121 y^{2}}{21} - \frac{1936 y}{105} + \frac{26927}{1680} = 0\)
\(\displaystyle \quad x^{2} + 4 x + \frac{121 y^{2}}{21} + \frac{484 y}{35} + \frac{261}{35} = 0\)
\(\displaystyle \quad x^{2} + 2 x + \frac{121 y^{2}}{21} - \frac{484 y}{35} + \frac{5037}{560} = 0\)
\(\displaystyle \quad x^{2} + 2 x + \frac{121 y^{2}}{21} - \frac{484 y}{15} + \frac{124}{3} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} - \frac{968 y}{35} + \frac{5037}{140} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} + \frac{242 y}{35} + \frac{681}{140} = 0\)
\(\displaystyle \quad x^{2} - 4 x + \frac{121 y^{2}}{21} + \frac{484 y}{35} + \frac{261}{35} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} + \frac{484 y}{105} - \frac{269}{336} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} + \frac{242 y}{105} - \frac{379}{105} = 0\)
\(\displaystyle \quad x^{2} - 2 x + \frac{121 y^{2}}{21} - \frac{3146 y}{105} + \frac{62543}{1680} = 0\)
Cada olho é uma elipse de eixo maior vertical com os valores de \(a\) e de \(c\) dados abaixo (veja o seu número \(n\) nesta lista).
Ache as equações gerais destas elipses.
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{1}{3}\ ,\quad c = \frac{1}{4}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{1}{3}\ ,\quad c = \frac{1}{4}\)
\(\displaystyle \quad a = \frac{2}{3}\ ,\quad c = \frac{1}{2}\)
\(\displaystyle \quad a = \frac{2}{3}\ ,\quad c = \frac{1}{2}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{4}{3}\ ,\quad c = 1\)
\(\displaystyle \quad a = \frac{1}{3}\ ,\quad c = \frac{1}{4}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{4}{3}\ ,\quad c = 1\)
\(\displaystyle \quad a = \frac{1}{3}\ ,\quad c = \frac{1}{4}\)
\(\displaystyle \quad a = \frac{2}{3}\ ,\quad c = \frac{1}{2}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{2}{3}\ ,\quad c = \frac{1}{2}\)
\(\displaystyle \quad a = \frac{4}{3}\ ,\quad c = 1\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{1}{3}\ ,\quad c = \frac{1}{4}\)
\(\displaystyle \quad a = \frac{2}{3}\ ,\quad c = \frac{1}{2}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{2}{3}\ ,\quad c = \frac{1}{2}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{1}{3}\ ,\quad c = \frac{1}{4}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{4}{3}\ ,\quad c = 1\)
\(\displaystyle \quad a = \frac{1}{3}\ ,\quad c = \frac{1}{4}\)
\(\displaystyle \quad a = \frac{4}{3}\ ,\quad c = 1\)
\(\displaystyle \quad a = \frac{1}{3}\ ,\quad c = \frac{1}{4}\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{4}{3}\ ,\quad c = 1\)
\(\displaystyle \quad a = \frac{1}{3}\ ,\quad c = \frac{1}{4}\)
\(\displaystyle \quad a = \frac{4}{3}\ ,\quad c = 1\)
\(\displaystyle \quad a = \frac{2}{3}\ ,\quad c = \frac{1}{2}\)
\(\displaystyle \quad a = \frac{2}{3}\ ,\quad c = \frac{1}{2}\)
\(\displaystyle \quad a = \frac{4}{3}\ ,\quad c = 1\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
\(\displaystyle \quad a = \frac{4}{3}\ ,\quad c = 1\)
\(\displaystyle \quad a = 1\ ,\quad c = \frac{3}{4}\)
A parte superior da boca é uma parábola com concavidade para cima, com as coordenadas do foco \(F\) e a distância focal \(p\) dadas abaixo (veja o seu número \(n\) nesta lista).
Ache a equação canônica desta parábola.
\(\displaystyle \quad F = \left( -1; \ -3\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( -1; \ - \frac{7}{3}\right)\ ,\quad p = \frac{1}{4}\)
\(\displaystyle \quad F = \left( 2; \ 1\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( 1; \ - \frac{4}{3}\right)\ ,\quad p = \frac{1}{4}\)
\(\displaystyle \quad F = \left( 2; \ \frac{1}{3}\right)\ ,\quad p = \frac{1}{2}\)
\(\displaystyle \quad F = \left( 1; \ \frac{1}{3}\right)\ ,\quad p = \frac{1}{2}\)
\(\displaystyle \quad F = \left( -2; \ -3\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( 2; \ \frac{2}{3}\right)\ ,\quad p = 1\)
\(\displaystyle \quad F = \left( -2; \ \frac{5}{3}\right)\ ,\quad p = \frac{1}{4}\)
\(\displaystyle \quad F = \left( -1; \ 1\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( 2; \ - \frac{1}{3}\right)\ ,\quad p = 1\)
\(\displaystyle \quad F = \left( 2; \ - \frac{7}{3}\right)\ ,\quad p = \frac{1}{4}\)
\(\displaystyle \quad F = \left( -2; \ \frac{1}{3}\right)\ ,\quad p = \frac{1}{2}\)
\(\displaystyle \quad F = \left( 1; \ 0\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( -1; \ 0\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( 1; \ - \frac{8}{3}\right)\ ,\quad p = \frac{1}{2}\)
\(\displaystyle \quad F = \left( 1; \ - \frac{1}{3}\right)\ ,\quad p = 1\)
\(\displaystyle \quad F = \left( 1; \ -3\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( 1; \ \frac{5}{3}\right)\ ,\quad p = \frac{1}{4}\)
\(\displaystyle \quad F = \left( -1; \ - \frac{5}{3}\right)\ ,\quad p = \frac{1}{2}\)
\(\displaystyle \quad F = \left( 2; \ 0\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( -1; \ -2\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( 2; \ -2\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( -1; \ \frac{1}{3}\right)\ ,\quad p = \frac{1}{2}\)
\(\displaystyle \quad F = \left( -2; \ -2\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( -2; \ 1\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( 2; \ \frac{5}{3}\right)\ ,\quad p = \frac{1}{4}\)
\(\displaystyle \quad F = \left( 2; \ -3\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( -2; \ - \frac{1}{3}\right)\ ,\quad p = 1\)
\(\displaystyle \quad F = \left( 1; \ \frac{2}{3}\right)\ ,\quad p = \frac{1}{4}\)
\(\displaystyle \quad F = \left( -1; \ - \frac{7}{3}\right)\ ,\quad p = 1\)
\(\displaystyle \quad F = \left( -2; \ - \frac{4}{3}\right)\ ,\quad p = \frac{1}{4}\)
\(\displaystyle \quad F = \left( -2; \ 0\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( -2; \ - \frac{10}{3}\right)\ ,\quad p = 1\)
\(\displaystyle \quad F = \left( -1; \ \frac{2}{3}\right)\ ,\quad p = \frac{1}{4}\)
\(\displaystyle \quad F = \left( -1; \ \frac{2}{3}\right)\ ,\quad p = 1\)
\(\displaystyle \quad F = \left( 2; \ \frac{4}{3}\right)\ ,\quad p = \frac{1}{2}\)
\(\displaystyle \quad F = \left( 2; \ - \frac{5}{3}\right)\ ,\quad p = \frac{1}{2}\)
\(\displaystyle \quad F = \left( 2; \ - \frac{10}{3}\right)\ ,\quad p = 1\)
\(\displaystyle \quad F = \left( 1; \ -2\right)\ ,\quad p = \frac{3}{4}\)
\(\displaystyle \quad F = \left( 1; \ - \frac{7}{3}\right)\ ,\quad p = 1\)
\(\displaystyle \quad F = \left( 1; \ 1\right)\ ,\quad p = \frac{3}{4}\)
A parte inferior da boca também é uma parábola, cuja equação geral é dada abaixo (veja o seu número \(n\) nesta lista).
Ache a equação canônica desta parábola.
\(\displaystyle \quad x^{2} + 2 x - \frac{27 y}{13} - \frac{217}{26} = 0\)
\(\displaystyle \quad x^{2} + 2 x - \frac{9 y}{13} - \frac{25}{26} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{27 y}{13} + \frac{77}{26} = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{9 y}{13} - \frac{7}{26} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{18 y}{13} + \frac{40}{13} = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{18 y}{13} + \frac{1}{13} = 0\)
\(\displaystyle \quad x^{2} + 4 x - \frac{27 y}{13} - \frac{139}{26} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{36 y}{13} + \frac{4}{13} = 0\)
\(\displaystyle \quad x^{2} + 4 x - \frac{9 y}{13} + \frac{125}{26} = 0\)
\(\displaystyle \quad x^{2} + 2 x - \frac{27 y}{13} - \frac{1}{26} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{36 y}{13} - \frac{32}{13} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{9 y}{13} + \frac{53}{26} = 0\)
\(\displaystyle \quad x^{2} + 4 x - \frac{18 y}{13} + \frac{40}{13} = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{27 y}{13} - \frac{55}{26} = 0\)
\(\displaystyle \quad x^{2} + 2 x - \frac{27 y}{13} - \frac{55}{26} = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{18 y}{13} - \frac{53}{13} = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{36 y}{13} - \frac{71}{13} = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{27 y}{13} - \frac{217}{26} = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{9 y}{13} + \frac{47}{26} = 0\)
\(\displaystyle \quad x^{2} + 2 x - \frac{18 y}{13} - \frac{35}{13} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{27 y}{13} + \frac{23}{26} = 0\)
\(\displaystyle \quad x^{2} + 2 x - \frac{27 y}{13} - \frac{163}{26} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{27 y}{13} - \frac{85}{26} = 0\)
\(\displaystyle \quad x^{2} + 2 x - \frac{18 y}{13} + \frac{1}{13} = 0\)
\(\displaystyle \quad x^{2} + 4 x - \frac{27 y}{13} - \frac{85}{26} = 0\)
\(\displaystyle \quad x^{2} + 4 x - \frac{27 y}{13} + \frac{77}{26} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{9 y}{13} + \frac{125}{26} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{27 y}{13} - \frac{139}{26} = 0\)
\(\displaystyle \quad x^{2} + 4 x - \frac{36 y}{13} - \frac{32}{13} = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{9 y}{13} + \frac{29}{26} = 0\)
\(\displaystyle \quad x^{2} + 2 x - \frac{36 y}{13} - 11 = 0\)
\(\displaystyle \quad x^{2} + 4 x - \frac{9 y}{13} + \frac{71}{26} = 0\)
\(\displaystyle \quad x^{2} + 4 x - \frac{27 y}{13} + \frac{23}{26} = 0\)
\(\displaystyle \quad x^{2} + 4 x - \frac{36 y}{13} - \frac{140}{13} = 0\)
\(\displaystyle \quad x^{2} + 2 x - \frac{9 y}{13} + \frac{29}{26} = 0\)
\(\displaystyle \quad x^{2} + 2 x - \frac{36 y}{13} - \frac{35}{13} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{18 y}{13} + \frac{58}{13} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{18 y}{13} + \frac{4}{13} = 0\)
\(\displaystyle \quad x^{2} - 4 x - \frac{36 y}{13} - \frac{140}{13} = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{27 y}{13} - \frac{163}{26} = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{36 y}{13} - 11 = 0\)
\(\displaystyle \quad x^{2} - 2 x - \frac{27 y}{13} - \frac{1}{26} = 0\)
Escreva uma inequação que represente a região da boca que está preenchida em vermelho na figura.
No Geogebra, entre a inequação e configure-a para que a área seja preenchida em vermelho.
As orelhas são os ramos de uma hipérbole de eixo real horizontal, de excentricidade \(11/10\), cujos vértices são os dois pontos do círculo do rosto que têm a coordenada \(y\) igual ao valor abaixo (veja o seu número \(n\) nesta lista).
\(\displaystyle \quad y = -2\)
\(\displaystyle \quad y = -2\)
\(\displaystyle \quad y = 2\)
\(\displaystyle \quad y = -1\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = -2\)
\(\displaystyle \quad y = 2\)
\(\displaystyle \quad y = 2\)
\(\displaystyle \quad y = 2\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = -2\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = -2\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = -2\)
\(\displaystyle \quad y = 2\)
\(\displaystyle \quad y = -1\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = -1\)
\(\displaystyle \quad y = -1\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = -1\)
\(\displaystyle \quad y = 2\)
\(\displaystyle \quad y = 2\)
\(\displaystyle \quad y = -2\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = -1\)
\(\displaystyle \quad y = -1\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = -2\)
\(\displaystyle \quad y = 1\)
\(\displaystyle \quad y = 2\)
\(\displaystyle \quad y = 2\)
\(\displaystyle \quad y = -1\)
\(\displaystyle \quad y = -2\)
\(\displaystyle \quad y = -1\)
\(\displaystyle \quad y = -1\)
\(\displaystyle \quad y = 2\)
Ache a equação canônica desta hipérbole.
Escreva uma inequação que represente a região das orelhas que está preenchida em rosa na figura. A distância \(d\) entre o centro do rosto e a borda vertical de cada orelha é a dada abaixo (veja o seu número \(n\) nesta lista).
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{8}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{8}\)
\(\displaystyle \quad d = \frac{9}{4}\)
\(\displaystyle \quad d = \frac{9}{4}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{2}\)
\(\displaystyle \quad d = \frac{9}{8}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{2}\)
\(\displaystyle \quad d = \frac{9}{8}\)
\(\displaystyle \quad d = \frac{9}{4}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{4}\)
\(\displaystyle \quad d = \frac{9}{2}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{8}\)
\(\displaystyle \quad d = \frac{9}{4}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{4}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{8}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{2}\)
\(\displaystyle \quad d = \frac{9}{8}\)
\(\displaystyle \quad d = \frac{9}{2}\)
\(\displaystyle \quad d = \frac{9}{8}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{2}\)
\(\displaystyle \quad d = \frac{9}{8}\)
\(\displaystyle \quad d = \frac{9}{2}\)
\(\displaystyle \quad d = \frac{9}{4}\)
\(\displaystyle \quad d = \frac{9}{4}\)
\(\displaystyle \quad d = \frac{9}{2}\)
\(\displaystyle \quad d = \frac{27}{8}\)
\(\displaystyle \quad d = \frac{9}{2}\)
\(\displaystyle \quad d = \frac{27}{8}\)
No Geogebra, entre a inequação e configure-a para que a área seja preenchida em rosa.
matrícula | n |
---|---|
119060029 | 4 |
122060003 | 8 |
122060004 | 37 |
122060005 | 22 |
122060006 | 32 |
122060007 | 17 |
122060008 | 36 |
122060009 | 35 |
122060010 | 41 |
122060011 | 18 |
122060012 | 7 |
122060013 | 28 |
122060014 | 27 |
122060015 | 33 |
122060016 | 3 |
122060017 | 6 |
122060018 | 15 |
122060019 | 31 |
122060020 | 26 |
122060021 | 29 |
122060022 | 19 |
122060023 | 25 |
122060028 | 42 |
122060029 | 1 |
122060030 | 13 |
122060031 | 16 |
122060033 | 20 |
122060034 | 24 |
122060035 | 34 |
122060036 | 21 |
122060038 | 9 |
122060040 | 5 |
122060041 | 40 |
215060056 | 2 |
220060041 | 14 |
221060040 | 12 |
221060047 | 39 |
622060024 | 10 |
622060025 | 30 |
622060026 | 23 |
622060027 | 38 |
822060037 | 11 |