2
$\begingroup$

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

$\endgroup$
3
  • 2
    $\begingroup$ do you know Lagrange multipliers? $\endgroup$
    – gt6989b
    Mar 13, 2019 at 3:00
  • $\begingroup$ yes @gt6989b.... $\endgroup$
    – jacky
    Mar 13, 2019 at 3:33
  • $\begingroup$ 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. $\endgroup$
    – farruhota
    Mar 13, 2019 at 5:21

2 Answers 2

7
$\begingroup$

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 $(3\sqrt {2})^2=18$

$\endgroup$
4
$\begingroup$

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

The equality occurs if $2xy=3$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .