Question

algúem consegue completar quadrado dessa equação?tentei mas está dando errado
$5 x^{2} +10 y^{2} -26x-32y-5=0$

Answer 1

Escrever a equação da cônica na forma reduzida (canônica), usando completamento de quadrados:

$5x^2+10y^2-26x-32y-5=0\\\\ (5x^2-26x)+(10y^2-32y)-5=0\\\\ (5x^2-26x)+2(5y^2-16y)-5=0\qquad\quad\mathbf{(i)}$

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Vamos completar os quadrados das expressões envolvendo $x$ e $y$ separadamente:

$5x^2-26x=\dfrac{1}{5}\cdot 5(5x^2-26x)\\\\\\ 5x^2-26x=\dfrac{1}{5}\,(25x^2-26\cdot 5x)\\\\\\ 5x^2-26x=\dfrac{1}{5}\,(25x^2-2\cdot 13\cdot 5x)\\\\\\ 5x^2-26x=\dfrac{1}{5}\,[(5x)^2-2\cdot 13\cdot 5x]$

Dentro dos colchetes, some e subtraia $13^2:$

$5x^2-26x=\dfrac{1}{5}\,[(5x)^2-2\cdot 13\cdot 5x+13^2-13^2]\\\\\\ 5x^2-26x=\dfrac{1}{5}\,[(5x-13)^2-13^2]\\\\\\ 5x^2-26x=\dfrac{1}{5}\left[\left(5\Big(x-\dfrac{13}{5}\Big)\right)^2-169\right]\\\\\\ 5x^2-26x=\dfrac{1}{5}\left[25\left(x-\dfrac{13}{5}\right)^2-169\right]$

$5x^2-26x=\dfrac{25}{5}\left(x-\dfrac{13}{5}\right)^2-\dfrac{169}{5}\\\\\\ 5x^2-26x=5\left(x-\dfrac{13}{5}\right)^2-\dfrac{169}{5}\qquad\quad\mathbf{(ii)}$

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Agora a expressão que envolve $y:$

$5y^2-16y=\dfrac{1}{5}\cdot 5(5y^2-16y)\\\\\\ 5y^2-16y=\dfrac{1}{5}\,(25y^2-16\cdot 5y)\\\\\\ 5y^2-16y=\dfrac{1}{5}\,(25y^2-2\cdot 8\cdot 5y)\\\\\\ 5y^2-16y=\dfrac{1}{5}\,[(5y)^2-2\cdot 8\cdot 5y]$

Dentro dos colchetes, some e subtraia $8^2:$

$5y^2-16y=\dfrac{1}{5}\,[(5y)^2-2\cdot 8\cdot 5y+8^2-8^2]\\\\\\ 5y^2-16y=\dfrac{1}{5}\,[(5y-8)^2-8^2]\\\\\\ 5y^2-16y=\dfrac{1}{5}\left[\left(5\Big(y-\dfrac{8}{5}\Big)\right)^2-64\right]\\\\\\ 5y^2-16y=\dfrac{1}{5}\left[25\left(y-\dfrac{8}{5}\right)^2-64\right]$

$5y^2-16y=\dfrac{25}{5}\left(y-\dfrac{8}{5}\right)^2-\dfrac{64}{5}\\\\\\ 5y^2-16y=5\left(y-\dfrac{8}{5}\right)^2-\dfrac{64}{5}$

Multiplique os dois lados por 2:

$2(5y^2-16y)=10\left(y-\dfrac{8}{5}\right)^2-\dfrac{128}{5} \qquad\quad\mathbf{(iii)}$

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

$\left[5\left(x-\dfrac{13}{5}\right)^2-\dfrac{169}{5}\right]+\left[10\left(y-\frac{8}{5}\right)^2-\dfrac{128}{5}\right]-5=0\\\\\\ 5\left(x-\dfrac{13}{5}\right)^2+10\left(y-\frac{8}{5}\right)^2-\dfrac{169}{5}-\dfrac{128}{5}-5=0\\\\\\ 5\left(x-\dfrac{13}{5}\right)^2+10\left(y-\frac{8}{5}\right)^2 +\dfrac{-169-128-25}{5}=0\\\\\\ 5\left(x-\dfrac{13}{5}\right)^2+10\left(y-\frac{8}{5}\right)^2-\dfrac{322}{5}=0\\\\\\ 5\left(x-\dfrac{13}{5}\right)^2+10\left(y-\frac{8}{5}\right)^2=\dfrac{322}{5}$

Multiplique os dois lados por $\dfrac{5}{322}:$

$\dfrac{25}{322}\left(x-\dfrac{13}{5}\right)^2+\dfrac{50}{322}\left(y-\frac{8}{5}\right)^2=1\\\\\\ \dfrac{\left(x-\frac{13}{5}\right)^2}{\frac{322}{25}}+\dfrac{\left(y-\frac{8}{5}\right)^2}{\frac{322}{50}}=1$

$\boxed{\begin{array}{c}\dfrac{\left(x-\frac{13}{5}\right)^2}{\frac{322}{25}}+\dfrac{\left(y-\frac{8}{5}\right)^2}{\frac{161}{25}}=1\end{array}}$

Esta é a equação de uma elipse deslocada, com centro no ponto $\left(\dfrac{13}{5},\,\dfrac{8}{5}\right),$ e semieixos medindo

$\left\{\!\begin{array}{l}a=\dfrac{\sqrt{322}}{5}\\\\ b=\dfrac{\sqrt{161}}{5} \end{array}\right.$

Bons estudos! :-)