# 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

• do you know Lagrange multipliers? Mar 13, 2019 at 3:00
• yes @gt6989b.... Mar 13, 2019 at 3:33
• note that the line starting "put..." is not true, because it is not $\cos^2 \alpha\color{red}+\sin^2 \alpha=1$. In other words, it is a hyperbola, not an ellipse. Mar 13, 2019 at 5:21

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