Question

Determine o centro, os focos e os vertices de hipérbole de equação 3x² - y² + 18x + 2y + 38 = 0​

Answer 1

✅ Após resolver os cálculos, concluímos que o centro, os focos e os vértices da referida hipérbole são, respectivamente:

                            $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf C(-3,\,1)\:\:\:}}\end{gathered} }$

            $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf F'(-3,\,-3)\:\:\:e\:\:\:F''(-3, \,5)\:\:\:}}\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf V'(-3,\,1 - 2\sqrt{3})\:\:\:e\:\:\:V''(-3,\,1 + 2\sqrt{3})\:\:\:}}\end{gathered} }$

Seja a hipérbole dada:

      $\Large\displaystyle\begin{gathered} 3x^{2} - y^{2} + 18x + 2y + 38 = 0\end{gathered}$

Para começar a solução, devemos completar os quadrados do polinômio que representa a equação geral da hipérbole tanto em "x" quanto em "y" e depois simplificar até chegar à uma das seguintes formas reduzida:

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

                                   ou

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

Observe que a equação "I" se refere à equação reduzida da hipérbole cujo eixo real é paralelo ao eixo das abscissas.

Já a equação "II" se refere à equação reduzida da hipérbole cujo eixo real é paralelo ao eixo das ordenadas.

Onde:

      $\Large\begin{cases} a = metade\:do\:eixo\:real\\b = metade\:do\:eixo\:imagin\acute{a}rio\\c = \sqrt{a^{2} + b^{2}}\end{cases}$

Completando os quadrados, temos:

               $\Large\displaystyle\begin{gathered} 3x^{2} - y^{2} + 18x + 2y + 38 = 0\end{gathered}$

                          $\Large\displaystyle\begin{gathered} 3x^{2} - y^{2} + 18x + 2y = -38\end{gathered}$

$\displaystyle\begin{gathered} 3\cdot\left[x^{2} + 6x + \bigg(\frac{6}{2}\bigg)^{2}\right] - \left[y^{2} - 2y + \bigg(\frac{-2}{2}\bigg)^{2}\right] = -38 + 3\cdot\bigg(\frac{6}{2}\bigg)^{2} - \bigg(\frac{-2}{2}\bigg)^{2}\end{gathered}$

$\Large\displaystyle\begin{gathered} 3\cdot\left[x^{2} + 6x + 9\right] - \left[y^{2} - 2y + 1\right] = -38 + 27 - 1\end{gathered}$

                         $\Large\displaystyle\begin{gathered} 3\cdot\left[x + 3\right]^{2} - \left[y - 1\right]^{2} = -12 \end{gathered}$      

                      $\Large\displaystyle\text{ \begin{gathered} \frac{3\cdot(x + 3)^{2}}{-12} - \frac{(y - 1)^{2}}{-12} = \frac{-12}{-12}\end{gathered} }$

                            $\Large\displaystyle\text{ \begin{gathered} \frac{(x + 3)^{2}}{-4} - \frac{(y - 1)^{2}}{-12}= 1\end{gathered} }$

                           $\Large\displaystyle\text{ \begin{gathered} \frac{(y - 1)^{2}}{12} - \frac{(x + 3)^{2}}{4} = 1\end{gathered} }$

• Calculando o valor de "a":  

                              $\Large\displaystyle\begin{gathered} a^{2} = 12 \Longrightarrow a = 2\sqrt{3}\end{gathered}$

• Calculando o valor de "b":

                                     $\Large\displaystyle\begin{gathered} b^{2} = 4 \Longrightarrow b = 2\end{gathered}$

• Calculando o valor de "c":

                              $\Large\displaystyle\begin{gathered} c = \sqrt{12 + 4} = \sqrt{16} = 4\end{gathered}$

• Calculando o centro da hipérbole:

                      $\Large\displaystyle\begin{gathered} C = (X_{C},Y_{C}) = (-3, 1)\end{gathered}$

• Calculando F':

                         $\Large\displaystyle\begin{gathered} F' = (X_{C}, Y_{C} - c) \end{gathered}$

                                $\Large\displaystyle\begin{gathered} = (-3, 1 - 4)\end{gathered}$

                                $\Large\displaystyle\begin{gathered} = (-3, -3)\end{gathered}$

                           $\Large\displaystyle\begin{gathered} \therefore\:\:F'(-3, -3)\end{gathered}$

• Calculando F'':

                          $\Large\displaystyle\begin{gathered} F' = (X_{C}, Y_{C} + c) \end{gathered}$

                                 $\Large\displaystyle\begin{gathered} = (-3, 1 + 4)\end{gathered}$

                                 $\Large\displaystyle\begin{gathered} = (-3, 5)\end{gathered}$

                             $\Large\displaystyle\begin{gathered} \therefore\:\:F''(-3, 5)\end{gathered}$

• Calculando V':

         $\Large\displaystyle\begin{gathered} V' = (X_{C}, Y_{C} - a) = (-3, 1 - 2\sqrt{3})\end{gathered}$

• Calculando V'':

         $\Large\displaystyle\begin{gathered} V' = (X_{C}, Y_{C} + a) = (-3, 1 + 2\sqrt{3})\end{gathered}$    

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

Saiba mais:

1. https://brainly.com.br/tarefa/38484156
2. https://brainly.com.br/tarefa/52646926
3. https://brainly.com.br/tarefa/53336195

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