Why do I always get a point on the circle

Question

Consider a hyperbola on the affine plane $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ for some positive real number $a,b$. Consider a circle $C: x^2 + y^2 = a^2$ that touches the two vertices of the hyperbola $H$. Fix a line $L:x=a$. Let $V = (-a,0)$ be the left vertex and let $P\in H$ be a point not on the $x$-axis. The intersection of the line $VP$ and the circle $C$ consists of two point (one being $V$) and we choose $P'$ to be the point that is not $V$. So we are now given the following fixed objects: $C,H,L,P,P'$.

My question is.. why does the following construction always end up in a point $Q'$ that is in the circle $C$

Let $Q$ be a point on the hyperbola (not $P$). Let $F=L\cap PQ$ and define $$Q' := FP' \cap VQ$$

I suppose I could just write down the analytic equations in order to prove this. But maybe there is a smarter way to argue this?

Answer 1

The construction is more general than is presented, and the result perhaps more surprising. I think there must be a projective proof here, but I'll just provide a brute-force analytical one. First, I'll restate the problem and change some notation.

Consider the conics with equations $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \qquad \frac{x^2}{a^2}+\frac{y^2}{c^2}=1$$ where $a$ is some non-zero real number, but $b$ and $c$ can be either real or imaginary (but non-zero), so that each corresponding equation can represent either an ellipse or a hyperbola. The figure takes $b$ imaginary and $c$ real. Let $V$ and $W$ be points $(0,-a)$ and $(0,a)$, common to both curves. Let $P$ and $Q$ be distinct points (and $\neq W$) on the common tangent at $W$, and define $p := |PW|$, $q := |QW|$ (where we can take these to be signed distances). Let $P_b$ and $Q_b$ be the points (other than $V$) where $\overleftrightarrow{VP}$ and $\overleftrightarrow{VQ}$ meet the "$b$"-conic; likewise, define $P_c$ and $Q_c$ on the "$c$"-conic. Then $M := \overleftrightarrow{P_bQ_b}\cap\overleftrightarrow{P_cQ_c}$ lies on that common tangent line; moreover, $$|MW| = \frac{pq}{p+q} = \frac{1}{\frac{1}{p}+\frac{1}{q}}=\frac12\cdot(\text{the harmonic mean of $p$ and $q$})$$ (Note that this value is independent of $b$ and $c$.)

Proof. Starting with $P = (a,p)$ and $Q=(a,q)$, we let Mathematica do all the symbol crunching to yield $$P_b = \left(a\,\frac{4 b^2 - p^2}{4 b^2 + p^2}, \frac{4 b^2 p}{4 b^2 + p^2}\right)$$ with $P_c$, $Q_b$, $Q_c$ derived by simply substituting $b\to c$ and/or $p\to q$. (Note that $b^2$ and/or $c^2$ could be negative. However, we never see $b$ or $c$ to the first power, so the real- or imaginary-ness of the values doesn't otherwise affect the calculations.) The lines in question are simply $$\begin{align} (4 b^2-pq)x + 2 a (p+q) y &= a ( 4 b^2 + p q ) \\ (4 c^2-pq)x + 2 a (p+q) y &= a ( 4 c^2 + p q ) \\ \end{align}$$ from which we readily determine the intersection, $M = \left(a, \frac{pq}{p+q}\right)$, as claimed. $\square$

The irrelevance of the specific parameters $b$ and $c$ ---and the reason to suspect a projective proof--- can be illustrated thusly: