Minimum value of $(x^2+y^2)^2$ if $x,y$ are real number such that $x^2+2xy-y^2=6$ Then find minimum value of $(x^2+y^2)^2$
what i try : $x^2+2xy+y^2-2y^2=6$ or $(x+y)^2-\bigg(\sqrt{2} y\bigg)^2=6$
put $\displaystyle (x+y)=\sqrt{6}\cos \alpha$ and $\displaystyle \sqrt{2}y=\sqrt{6}\sin \alpha$
$\displaystyle x=\sqrt{6}\cos \alpha-\sqrt{3}\sin \alpha$ and $\displaystyle y =\sqrt{3}\sin \alpha$
$\displaystyle x^2+y^2=3\bigg[\bigg(\sqrt{2}\cos \alpha-\sin \alpha\bigg)^2+\sin^2\alpha\bigg)\bigg]$
$\displaystyle x^2+y^2=3\bigg[2\cos^2\alpha+\sin^2\alpha-2\sqrt{2}\cos \alpha\sin \alpha+\sin^2\alpha\bigg]$
$\displaystyle x^2+y^2=3\bigg(2-\sqrt{2}\sin 2\alpha\bigg)\geq 3(2-\sqrt{2})$
but answer is $\sqrt{18}$ How do i solve it Help me please 

$$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=(6-2xy)^2+4x^2y^2=2(2xy-3)^2+18\ge18$$
The equality occurs if $2xy=3$
