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If $x^{2} + x y + y^{2} = a^{2}$ , then $y_{2} =$

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$\frac{- 6 a^{2}}{{(x - 2 y)}^{3}}$

$\frac{- 6 a^{2}}{{(x + 2 y)}^{3}}$

$\frac{- 3 a^{2}}{{(x + 2 y)}^{3}}$

$\frac{3 a^{2}}{{(x - 2 y)}^{3}}$

answer is A.

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Detailed Solution

$x^{2} + x y + y^{2} = a^{2}$

${(x + y)}^{2} = a^{2} + x y$

$2 (x + y) (1 + y_{1}) = x y_{1} + y$

$2 x + 2 x y_{1} + 2 y + 2 y y_{1} = x y_{1} + y$

$x y_{1} + y + 2 x + 2 y y_{1} = 0$

$y_{1} = \frac{- (2 x + y)}{x + 2 y}$

$y_{2} = - [\frac{(x + 2 y) (2 + y_{1}) - (2 x + y) (1 + 2 y_{1})}{{(x + 2 y)}^{2}}]$

$y_{2} = \frac{3 x y_{1} - 3 y}{{(x + 2 y)}^{2}}$

$y_{2} = \frac{- 3 x \frac{(2 x + y)}{x + 2 y} - 3 y}{{(x + 2 y)}^{2}}$

$y_{2} = \frac{- 6 (x^{2} + x y + y^{2})}{{(x + 2 y)}^{3}}$

$y_{2} = \frac{- 6 a^{2}}{{(x + 2 y)}^{3}}$

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