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.
Best Answer
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
$$x^2+y^2 = 20 \left(\dfrac{x^2}{16}+\dfrac{y^2}{4}\right)^2$$