If x and y are real number such that $x^{2} + 2 x y - y^{2} = 6,$ find the minimum value of ${(x^{2} + y^{2})}^{2} .$

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Solution

$x^{2} + 2 x y - y^{2} = 6$
$2 x + 2 x \frac{d y}{d x} - 2 y + 2 y = 6 \frac{d y}{d x} = \frac{6 - 2 x}{2 x} \frac{d y}{d x} = 0 x = 3 9 + 6 y - y^{2} = 6 y m i n v a l u e = 3 - 2 \sqrt{3} {[3^{2} + {(3 - 2 \sqrt{3})}^{2}]}^{2} = 84.92$

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