I was doing an excercise, then I got stuck at ths question,
Find the locus of the mid point of the chord of contact of the tangents to the ellipse $x^2+4y^2 = 16$ which are at right angles. I tried to find the equation of the director circle and use it's properties but it didn't help. Then I thought to somehow get two relations between the mid point's coordinates and the slope of the chord of contact but I could find only one. $\alpha+4m\beta= 0$. Where $ m $ is the slope of chord of contact and $\alpha $ , $\beta$ are the cordinates of the mid point of the chord of contact.
You have already made the right connection to the Director Circle. You can finish in the following way:
Let the midpoint be $(h,k)$. Then the equation of the chord using Joachimstahl notation is $$T=S_1: \dfrac{hx}{16}+\dfrac{ky}{4} = \dfrac{h^2}{16}+\dfrac{k^2}{4}$$
Let the pair of orthogonal tangents be drawn from $P(x_1,y_1)$. Then the chord of contact is $$\dfrac{xx_1}{16}+\dfrac{yy_1}{4} = 1$$
Since both represent the same straight line, we have
$$\dfrac{x_1}{h} = \dfrac{y_1}{k} = \dfrac{1}{\dfrac{h^2}{16}+\dfrac{k^2}{4}}$$
Since $P$ lies on the Director Circle $x^2+y^2 = 20$, we obtain the locus as
$$x^2+y^2 = 20 \left(\dfrac{x^2}{16}+\dfrac{y^2}{4}\right)^2$$