Question

URGENTE
Alguém completa o quadrado dessa equação?
$-8 x^{2} +2 y^{2} + \frac{8 \sqrt{10} }{5} x- \frac{14 \sqrt{10} }{5} y+9=0$

Answer 1

Reescrever a equação da cônica na forma reduzida, via completamento de quadrados:

     $-8x^2+2y^2+\dfrac{8\sqrt{10}}{5}\,x-\dfrac{14\sqrt{10}}{5}\,y+9=0\\\\\\ \bigg(-8x^2+\dfrac{8\sqrt{10}}{5}\,x\bigg)+\bigg(2y^2-\dfrac{14\sqrt{10}}{5}\,y\bigg)+9=0\qquad\mathbf{(i)}$


Para facilitar, vamos completar os quadrados das expressões entre parênteses separadamente.


     •   Expressão que envolve $x:$

     $-8x^2+\dfrac{8\sqrt{10}}{5}\,x=-8\cdot \bigg[x^2-\dfrac{\sqrt{10}}{5}\,x\bigg]\\\\\\ -8x^2+\dfrac{8\sqrt{10}}{5}\,x=-8\cdot \bigg[x^2-2\cdot x\cdot \dfrac{\sqrt{10}}{10}\bigg]$


Dentro dos colchetes do lado direito, some e subtraia $\dfrac{1}{10}=\bigg(\dfrac{\sqrt{10}}{10}\bigg)^2:$

     $-8x^2+\dfrac{8\sqrt{10}}{5}\,x=-8\cdot \bigg[x^2-2\cdot x\cdot \dfrac{\sqrt{10}}{10}+\dfrac{1}{10}-\dfrac{1}{10}\bigg]\\\\\\ -8x^2+\dfrac{8\sqrt{10}}{5}\,x=-8\cdot \bigg[x^2-2\cdot x\cdot \dfrac{\sqrt{10}}{10}+\bigg(\dfrac{\sqrt{10}}{10}\bigg)^2-\dfrac{1}{10}\bigg]\\\\\\ -8x^2+\dfrac{8\sqrt{10}}{5}\,x=-8\cdot \bigg[\bigg(x-\dfrac{\sqrt{10}}{10}\bigg)^2-\dfrac{1}{10}\bigg]\\\\\\ -8x^2+\dfrac{8\sqrt{10}}{5}\,x=-8\bigg(x-\dfrac{\sqrt{10}}{10}\bigg)^2+\dfrac{4}{5}\qquad\mathbf{(ii)}$


     •   Expressão que envolve $y:$

     $2y^2-\dfrac{14\sqrt{10}}{5}\,y=2\cdot \bigg[y^2-\dfrac{7\sqrt{10}}{5}\,y\bigg]\\\\\\ 2y^2-\dfrac{14\sqrt{10}}{5}\,y=2\cdot \bigg[y^2-2\cdot y\cdot \dfrac{7\sqrt{10}}{10}\bigg]$


Dentro dos colchetes do lado direito, some e subtraia $\dfrac{49}{10}=\bigg(\dfrac{7\sqrt{10}}{10}\bigg)^2:$

     $2y^2-\dfrac{14\sqrt{10}}{5}\,y=2\cdot \bigg[y^2-2\cdot y\cdot \dfrac{7\sqrt{10}}{10}+\dfrac{49}{10}-\dfrac{49}{10}\bigg]\\\\\\ 2y^2-\dfrac{14\sqrt{10}}{5}\,y=2\cdot \bigg[y^2-2\cdot y\cdot \dfrac{7\sqrt{10}}{10}+\bigg(\dfrac{7\sqrt{10}}{10}\bigg)^2-\dfrac{49}{10}\bigg]\\\\\\ 2y^2-\dfrac{14\sqrt{10}}{5}\,y=2\cdot \bigg[\bigg(y-\dfrac{7\sqrt{10}}{10}\bigg)^2-\dfrac{49}{10}\bigg]\\\\\\ 2y^2-\dfrac{14\sqrt{10}}{5}\,y=2\bigg(y-\dfrac{7\sqrt{10}}{10}\bigg)^2-\dfrac{49}{5}\qquad\mathbf{(iii)}$


Substituindo $\mathbf{(ii)}$ e $\mathbf{(iii)}$ na equação $\mathbf{(i)}$ da cônica, temos

     $\bigg[-8\bigg(x-\dfrac{\sqrt{10}}{10}\bigg)^2+\dfrac{4}{5}\bigg]+\bigg[2\bigg(y-\dfrac{7\sqrt{10}}{10}\bigg)^2-\dfrac{49}{5}\bigg]+9=0\\\\\\ -8\bigg(x-\dfrac{\sqrt{10}}{10}\bigg)^2+2\bigg(y-\dfrac{7\sqrt{10}}{10}\bigg)^2+\dfrac{4}{5}-\dfrac{49}{5}+9=0\\\\\\ -8\bigg(x-\dfrac{\sqrt{10}}{10}\bigg)^2+2\bigg(y-\dfrac{7\sqrt{10}}{10}\bigg)^2+\dfrac{4-49+45}{5}=0\\\\\\ -8\bigg(x-\dfrac{\sqrt{10}}{10}\bigg)^2+2\bigg(y-\dfrac{7\sqrt{10}}{10}\bigg)^2+0=0$


Divida os dois lados por $-8:$

     $\dfrac{(x-\frac{\sqrt{10}}{10})^2}{1}-\dfrac{(y-\frac{7\sqrt{10}}{10})^2}{4}=0\\\\\\ \dfrac{(x-\frac{\sqrt{10}}{10})^2}{1^2}-\dfrac{(y-\frac{7\sqrt{10}}{10})^2}{2^2}=0$


No lado esquerdo temos uma diferença entre quadrados igual a zero.

O lado esquerdo pode ser fatorado usando produtos notáveis (produto da diferença pela soma):

     $\bigg(\dfrac{x-\frac{\sqrt{10}}{10}}{1}-\dfrac{y-\frac{7\sqrt{10}}{10}}{2}\bigg)\cdot \bigg(\dfrac{x-\frac{\sqrt{10}}{10}}{1}+\dfrac{y-\frac{7\sqrt{10}}{10}}{2}\bigg)=0$


A equação dada representa duas retas concorrentes no plano.


Bons estudos! :-)