Question

Determine os focos da hipérbole

Answer 1

✅ Após resolver os cálculos, concluímos que os focos da hipérbole são, respectivamente:

 $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf F' = (1 - \sqrt{3}, -1)\:\:\:e\:\:\:F'' = (1 + \sqrt{3}, -1)\:\:\:}}\end{gathered} }$

Seja a equação geral da hipérbole:

                            $\Large\displaystyle\begin{gathered} x^{2} - 2y^{2} - 2x - 4y = 3\end{gathered}$

Para começar a solução, devemos completar os quadrados do polinômio em "x" e do polinômio em "y", simplificar e determinar a equação reduzida da mesma sob a seguinte fórmula:

                           $\Large\displaystyle\text{ \begin{gathered} \frac{(x - x_{o})^{2}}{a^{2}} - \frac{(y - y_{o})^{2}}{b^{2}} = 1\end{gathered} }$

Onde:

                         $\Large\begin{cases} a = \textrm{metade eixo real}\\b = \textrm{metade eixo imaginario}\\c = \sqrt{a^{2} + b^{2}}\end{cases}$

Então, temos:

                              $\Large\displaystyle\begin{gathered} x^{2} - 2y^{2} - 2x - 4y = 3\end{gathered}$

        $\displaystyle\begin{gathered} x^{2} - 2x + \bigg(\frac{-2}{2}\bigg)^{2} - 2\bigg(y^{2} + 2y + \bigg(\frac{2}{2}\bigg)^{2}\bigg) = 3 + \bigg(\frac{-2}{2}\bigg)^{2} - 2\cdot\bigg(\frac{2}{2}\bigg)^{2}\end{gathered}$

   $\Large\displaystyle\begin{gathered} (x^{2} - 2x + 1) - 2\cdot(y^{2} + 2y + 1) = 3 + 1 - 2\end{gathered}$

                          $\Large\displaystyle\begin{gathered} (x - 1)^{2} - 2\cdot(y + 1)^{2} = 2\end{gathered}$

                        $\Large\displaystyle\text{ \begin{gathered} \frac{(x - 1)^{2}}{2} - \frac{{\!\diagup\!\!\!\!2}\cdot(y + 1)^{2}}{\!\diagup\!\!\!\!2} = \frac{2}{2}\end{gathered} }$

                               $\Large\displaystyle\text{ \begin{gathered} \frac{(x - 1)^{2}}{2} - \frac{(y + 1)^{2}}{1} = 1\end{gathered} }$

Agora podemos recuperar as principais componentes da hipérbole que são:

                             $\Large\displaystyle\begin{gathered} O= (x_{o}, y_{o}) = (1, -1)\end{gathered}$

                                $\Large\displaystyle\begin{gathered} a^{2} = 2 \Longrightarrow a = \sqrt{2}\end{gathered}$

                             $\Large\displaystyle\begin{gathered} b^{2} = 1 \Longrightarrow b = \sqrt{1} = 1\end{gathered}$

     $\Large\displaystyle\text{ \begin{gathered} c = \sqrt{a^{2} + b^{2}} = \sqrt{(\sqrt{2})^{2} + 1^{2}} = \sqrt{2 + 1} = \sqrt{3}\end{gathered} }$

                   $\Large\displaystyle\begin{gathered} F' = (x_{o} - c, y_{o}) = (1 - \sqrt{3}, -1) \end{gathered}$

                   $\Large\displaystyle\begin{gathered} F'' = (x_{o} + c, y_{o}) = (1 + \sqrt{3}, -1)\end{gathered}$

✅ Portanto, os focos da hipérbole são:

           $\Large\displaystyle\begin{gathered} F' = (1 - \sqrt{3}, -1)\:\:\:e\:\:\:F'' = (1 + \sqrt{3}, -1)\end{gathered}$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

Saiba mais:

1. https://brainly.com.br/tarefa/38484156
2. https://brainly.com.br/tarefa/52646926

$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$