find a point on ellipse closest to origin Find the points on the ellipse $2x^2 + 4xy + 5y^2 = 30$ closest and farthest from origin. How to do this problem? I know how to find a closest point if $z = f(x,y)$ is given, however, this is 2 dimensional.

Another way is use Rotation of axes, to eliminate the $xy$ term.
Here $\cot 2\theta=\frac{2-5}4=-\frac 3 4$
$\frac {\cos 2\theta}3=\frac{\sin 2\theta}{-4}=\frac 1{\pm 5}$ (Using squaring & adding)
If $\cos 2\theta=\frac 3 5,\sin 2\theta=-\frac 4 5$
Using $\cos 2\theta=2\cos^2\theta-1=1-2\sin^2\theta$, we get $\sin^2\theta=\frac 1 5$ and $\cos^2\theta=\frac 4 5\implies \cos\theta=\pm\frac 2{\sqrt 5},\sin\theta=∓\frac 1{\sqrt 5}$ as $\sin 2\theta<0$, the sign of $\cos\theta,\sin\theta$ will be opposite.
The equation of the ellipse in the rotated axes becomes, $$\frac{x'^2}{30}+\frac{y'^2}{5}=1$$
Now in orthogonal transformation, the distance between two points is one of the invariants and in rotation, the origin remains unchanged and any point of this curve can be P$(\sqrt{30}\cos \phi,\sqrt5 \sin \phi)$.
The distance of P from the origin is $\sqrt{30\cos^2\phi+5 \sin^2 \phi}$
$=\sqrt{30-25\sin^2 \phi}$ 
The distance will be minimum $(=\sqrt 5)$ if $\sin^2 \phi=1$ i.e., from $(0,\pm\sqrt 5)$ in the rotated co-ordinates.
The distance will be maximum $(=\sqrt {30})$ if $\sin^2 \phi=0$ i.e., from $(\pm\sqrt {30},0)$ in the rotated co-ordinates.
Now use $x=x'\cos\theta-y'\sin\theta$ and $y=x'\sin\theta+y'\cos\theta$.
We have already derived, $\cos\theta=\pm\frac 2{\sqrt 5},\sin\theta=∓\frac 1{\sqrt 5}$ 
For the minimum distance, if $\cos\theta=\frac 2{\sqrt 5},\sin\theta=-\frac 1{\sqrt 5}$ 
If $y'=\sqrt 5,x=-\sqrt 5(-\frac 1{\sqrt 5})=1$ and $y=\sqrt5(\frac 2{\sqrt 5})=2$
Similarly, if $y'=-\sqrt 5, (x,y)$ will be $(-1,-2)$
The maximum case can be handled in the same way.
