I'm to determine the distance to origo from: $g(x,y)=13x^2+13y^2+10xy-72=0$.
I put: $f(x,y)=x^2+y^2$ for distance squared (easier to work with).
Using: det$\begin{vmatrix} \nabla f \\ \nabla g \end{vmatrix}$. I get $x=y$ and the stationary points $\pm(\sqrt2,\sqrt2)$.
How do I proceed from here?
I managed to get the correct answer by thinking about the constitution of a ellipse. Transforming my points $\pm(\sqrt2,\sqrt2)$ to polar coordinates and assuming that my next set of stationary points will be $\frac{\pi}{2}$ away.
Solving $a = \frac{\pi}{4}$ my other stationary points most be on $b=\frac{3\pi}{4}+k\pi, k=0,1$.
I then get the line $y=-x$ that goes through $b$.
Finally solving for x in g(x,y) and putting this = -y (for line intersections) gives me the correct answers $(\frac{3}{\sqrt2},\frac{-3}{\sqrt2})$ and $(\frac{-3}{\sqrt2},\frac{3}{\sqrt2})$.
However there most be an easier way to get this answer. Any tips?
Actually, we should have gotten all four points in the first round. The gradients are $$\nabla f = (2x,2y)\text{ , }\nabla g = (26x+10y,10x+26y)$$ Those are parallel when $(y,x)=\lambda(x,y)$ for some $\lambda$, or equivalently $x^2=y^2$. That's the union of two lines $y=x$ and $y=-x$, and our critical points come when they intersect the ellipse.
If $y=x$, the ellipse equation becomes $13x^2+13x^2+10x^2-72=0$, or $36x^2=72$ with solutions $x=\pm \sqrt{2}$. The two points we get from this are $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$. They are each at distance $2$ from the origin.
If $y=-x$, the ellipse equation becomes $13x^2+13x^2-10x^2-72=0$, or $16x^2=72$ with solutions $x=\pm\frac32\sqrt{2}$. The two points we get from this are $(\frac32\sqrt{2},-\frac32\sqrt{2})$ and $(-\frac32\sqrt{2},\frac32\sqrt{2})$. They are each at distance $3$ from the origin.
So then, with all critical points found, the minimum distance $2$ comes at $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$ while the maximum distance $3$ comes at $(\frac32\sqrt{2},-\frac32\sqrt{2})$ and $(-\frac32\sqrt{2},\frac32\sqrt{2})$.