Proving a relation between the ordinates of point of contact of a line on a circle

Question

If the tangents at $(h,k)$ to the ellipse $x²/a² + y²/b² = 1$ , cuts the auxillary circle in the point whose ordinates are $p$ and $q$ , then show that $1/p + 1/q = 2/k$ .

I found this question while I was solving questions on ellipses.

These may help in understanding the question better:

Tangent : The line which touches a curve only at one point is called a tangent.

Auxillary circle : the circle passing through the vertices on the major axis of an ellipse and whose diameter is equal to major axis of the ellipse is called its auxillary circle.

Now, I did not understand how to proceed. I just tried to write the equation of the tangent at $(h,k)$ and then solve it with the equation of the circle. But I don't understand why I am only supposed to prove it using mathematical formulae and equations. I want to know whether there is a good way to prove it using geometry.

Use of trigonometry, differentiations, integrations are allowed wherever necessary.

Answer 1

If the line through a point $C$ of an ellipse, perpendicular to the major axis, meets the auxiliary circle at $Q$, then the tangent to the ellipse at $C$ and the tangent to the circle at $Q$ meet the major axis at the same point $P$. This nice theorem is obvious if you think of the auxiliary circle as the ellipse stretched perpendicularly to the major axis, and allows to rephrase what you want to prove in a different way, with no connection to the ellipse:

Let $AB$ be a chord of a half-circle with diameter $RS$, line $AB$ meeting line $RS$ at $P$, and let line $PQ$ be tangent to the half-circle at $Q$. If $A'$, $B'$, $C'$ are the perpendicular projections of $A$, $B$, $Q$ on $RS$, and $C$ is the intersection between $QC'$ and $AB$, then: $${1\over AA'}+{1\over BB'}={2\over CC'}.$$

To prove that, note first of all that we can rewrite the above equality as $$ \tag{1} {CC'\over AA'}+{CC'\over BB'}=2. $$

On the other hand, if $O$ is the center of the circle, from similarity inside right triangle $PQO$ and power of point $P$ we have: $$ \tag{2} PQ^2=PC'\cdot PO=PA\cdot PB. $$ But $$ {AA'\over PA}={BB'\over PB}=\sin\alpha \quad\text{and}\quad {CC'\over PC'}=\tan\alpha, $$ where $\alpha=\angle APR$ (see figure). Hence equation $(2)$ can be rewritten as $$ CC'\cdot PO={AA'\cdot BB'\over\sin\alpha\cos\alpha} $$ whence we obtain, if $M$ is the midpoint of $AB$ and $M'$ its projection on $RS$: $$ {CC'\over AA'}+{CC'\over BB'}= {BB'+ AA'\over PO\sin\alpha\cos\alpha}= {BB'+ AA'\over MM'}=2, $$ as it was to be proved.