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 

I think a fun way to do the problem similar to what you've done is by using polar co-ordinates. Sub $x=r \cos \theta$ and $y=r \sin \theta$. Then you want to minimise $(x^2+y^2)^2=r^4$. Note that the constraint simplifies massively.
$$ x^2+2xy-y^2=6$$
$$r^2( \cos ^2 \theta +2 \sin \theta \cos \theta -\sin ^2 \theta )=6$$
$$r^2(\cos 2\theta +\sin 2\theta )=6$$
$$r^2 \sqrt{2} \sin ( 2\theta +\frac{\pi}{4})=6$$
$$r^2=3 \sqrt{2} \csc( 2\theta +\frac{\pi}{4})$$
Hence the minimum of $r^4$ is $(3\sqrt {2})^2=18$ 
