Question #3dd82

Oct 20, 2017

${x}^{2} + 4 {y}^{2} - 4 x + 4 y + 5 = 0$

$\Rightarrow {x}^{2} - 4 x + 4 + 4 {y}^{2} + 4 y + 1 = 0$

$\Rightarrow {\left(x - 2\right)}^{2} + {\left(2 y + 1\right)}^{2} = 0$

$\Rightarrow {\left(x - 2\right)}^{2} = 0 \mathmr{and} {\left(2 y + 1\right)}^{2} = 0$

$\Rightarrow x = 2 \mathmr{and} y = - \frac{1}{2}$

Putting these values after solving following equations :-

Given $\frac{{x}^{4} - {y}^{4}}{2 {x}^{2} + x y - {y}^{2}} . \frac{2 x - y}{x y - {y}^{2}} \div {\left({x}^{2} + {y}^{2} / y\right)}^{2}$

$\Rightarrow \frac{{\left({x}^{2}\right)}^{2} - {\left({y}^{2}\right)}^{2}}{2 {x}^{2} + 2 x y - x y - {y}^{2}} . \frac{2 x - y}{y \left(x - y\right)} \div {\left({x}^{2} + y\right)}^{2}$

$\Rightarrow \frac{\left({x}^{2} + {y}^{2}\right) \left({x}^{2} - {y}^{2}\right)}{2 x \left(x + y\right) - y \left(x + y\right)} . \frac{2 x - y}{y \left(x - y\right)} . \frac{1}{{x}^{2} + y} ^ 2$

$\Rightarrow \frac{\left({x}^{2} + {y}^{2}\right) \left(x + y\right) \left(x - y\right)}{\left(2 x - y\right) \left(x + y\right)} . \frac{2 x - y}{y \left(x - y\right)} . \frac{1}{{x}^{2} + y} ^ 2$

$\Rightarrow \frac{{x}^{2} + {y}^{2}}{y {\left({x}^{2} + y\right)}^{2}}$

$\Rightarrow \frac{{2}^{2} + {\left(- \frac{1}{2}\right)}^{2}}{\left(- \frac{1}{2}\right) {\left[{2}^{2} + \left(- \frac{1}{2}\right)\right]}^{2}}$

$\Rightarrow \frac{4 + \frac{1}{4}}{\left(- \frac{1}{2}\right) {\left\{4 - \frac{1}{2}\right\}}^{2}}$

$\Rightarrow \frac{\frac{17}{4}}{\left(- \frac{1}{2}\right) {\left(\frac{7}{2}\right)}^{2}}$

$\Rightarrow - \frac{17}{4.} \frac{2}{1.} \frac{4}{49}$

$\Rightarrow - \frac{34}{49}$