Question on Tangent

Question

Suppose that C is any circle concentric with the ellipse E: x^2/16+ y^2/4 =1 . Let A is a point on E and B is a point on C such that AB is tangent to both E and C. The maximum Length of AB is?

Answer 1

Let C be $\frac{x^2}{r^2}+\frac{y^2}{r^2}=1, 2\leq r\leq 4$. The dual conics to $E$ and $C$ are $16X^2+4Y^2=1$ and $r^2X^2+r^2Y^2=1$, respectively. They intersect in $(X,Y)=(\pm\frac{\sqrt{r^2-4}}{2\sqrt{3}r},\pm\frac{\sqrt{16-r^2}}{2\sqrt{3}r})$, that is the lines $\pm\frac{\sqrt{r^2-4}}{2\sqrt{3}r} x\pm\frac{\sqrt{16-r^2}}{2\sqrt{3}r} y+1=0$ are the common tangents, where the $\pm$s are taken independently. For more on the intersect-the-dual-conics approach to common tangents see this answer.

The length of the four segments are the same by symmetry so let's focus on the one in the first quadrant. The tangent points $B$ and $A$ are $(\frac{ r\sqrt{r^2-4}}{2\sqrt{3}}, \frac{r\sqrt{16-r^2}}{2\sqrt{3}})$ and $(\frac{8\sqrt{r^2-4}}{\sqrt{3}r},\frac{2\sqrt{16-r^2}}{\sqrt{3}r})$. So the length of $AB$ reduces to $$\sqrt{\frac{-r^4+20r^2-64}{r^2}}.$$ Now it's a calculus exercise to show that the maximum is $2$ for $r=\sqrt{8}$.