The maximum possible value of $ (xv - yu)^2 $ over the surface ... The maximum possible value of $ (xv - yu)^2 $ over the surface given by the equations 
$ x^2 + y^2 = 4 $ and $ u^2 + v^2 = 9 $ is :
I solved it and my answer comes out to be $9$ but the correct answer is $36$.
Here is what I did :
Put $ x = r_1\cos\theta$ and   $y = r_1\sin\theta $
$ u = r_2\sin\theta $ and $ v = r_2\cos\theta$
Using $ x^2 + y^2 = 4 $ and $ u^2 + v^2 = 9 $
we get $ r_1 = \sqrt 2 $ and $ r_2 = \frac 3{\sqrt 2} $
Now $ (xv - yu)^2 = (r_1r_2)^2(\cos^2\theta - \sin^2\theta)^2 $
$= 9(1-2\sin^2\theta)^2$
Now this will be maximum when $ (1- 2\sin^2\theta)^2 $ is maximum which is at $ \sin^2\theta = 0,1 $ so the maximum value of the expression comes out to be $9$.
Where did I go wrong here? Or am I correct?

Another way :
$$(xv-yu)^2=(x^2+y^2)(u^2+v^2)-(xu+yv)^2=36-(xu+yv)^2\le 36.$$
The equality is attained when $(x,y,u,v)=\left(\frac{4\sqrt 2}{3},-\frac 23,1,2\sqrt 2\right).$
P.S. When $(x,y,u,v)=\left(\frac{4\sqrt 2}{3},-\frac 23,1,2\sqrt 2\right)$, the followings hold :
$$xu+yv=0,\ \ \ x^2+y^2=4,\ \ \ u^2+v^2=9.$$
