# Locus of the midpoint of the chord of contact of orthogonal tangents.

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.

• Since lines, ellipses, tangents, and midpoints are affine concepts, it is enough to look at a circle inscribed in a square. Dec 7 '17 at 20:02
• Angle isn’t an affine concept, though.
– amd
Dec 7 '17 at 20:41

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

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