What is the locus of the midpoints of intercepts of tangents to the ellipse ${x^2\over a^2}+{y^2\over b^2}=1$ , cut off by its director circle?

Question

I used $T=0$ to get equation for tangent, but I am guessing I need one more equation for coefficient comparison. I also can't understand how to get to the coordinates of the points where the tangent intersects the circle, in order to get an expression for the mid-point.

Answer 1

Let $c=\sqrt{a^2+b^2}$, then the director circle is

$$x^2+y^2=c^2 \tag{1}$$

Let $c(\cos \theta,\sin \theta)$ and $c(\cos \phi,\sin \phi)$ be the points on $(1)$, the equation of the chord is

$$x\cos \frac{\theta+\phi}{2}+y\sin \frac{\theta+\phi}{2} =c\cos \frac{\theta-\phi}{2} \tag{2}$$

Let $(x',y')$ be a point on ellipse, then the tangent on it should be

$$\frac{x'x}{a^2}+\frac{y'y}{b^2}=1 \tag{3}$$

If $(2)$ is the polar of $(x',y')$ with respect to the ellipse, then $(2)$ and $(3)$ are equivalent.

Hence,

$$(x',y')= \frac{ \left( a^2\cos \frac{\theta+\phi}{2},b^2\sin \frac{\theta+\phi}{2} \right)} {c\cos \frac{\theta-\phi}{2}}$$

Note that the required mid-point is

$$(u,v)= c\cos \frac{\theta-\phi}{2} \left( \cos \frac{\theta+\phi}{2}, \sin \frac{\theta+\phi}{2} \right) $$

Now,

\begin{align} u^2+v^2 &= c^2 \cos^2 \frac{\theta-\phi}{2} \\ (x',y') &= \left( \frac{a^2u}{u^2+v^2}, \frac{b^2v}{u^2+v^2} \right) \\ 1 &= \frac{x'^2}{a^2}+\frac{y'^2}{b^2} \\ &= \frac{a^2u^2}{(u^2+v^2)^2}+\frac{b^2v^2}{(u^2+v^2)^2} \\ (u^2+v^2)^2 &= a^2 u^2+b^2 v^2 \end{align}

Therefore the required locus is

$$\fbox{$(x^2+y^2)^2=a^2 x^2+b^2 y^2$}$$

This locus is also the perpendicular foot of the tangent from the origin.