Conics property proof

Question

Prove this general Conics property of focal rays product as shown ( $a$ major axis, $b$ minor axis and focal ray makes angle $\psi$ with tangent) :

$$ r_1 \cdot r_2 = r (2 a -r) = b^2/\sin^2\psi $$

Answer 1

If we apply the Law of Cosines to your triangle, we find \begin{align*} (2c)^2 &= r_1^2 + r_2^2 - 2r_1r_2 \cos(\pi - 2 \psi) \\ 4c^2 &= r_1^2 +r_2^2 +2 r_1 r_2 \cos 2\psi \\ 4c^2 &= r_1^2 + r_2 ^2 +2r_1r_2(1-2 \sin ^2 \psi) \\ 4c^2 &= (r_1+r_2)^2 - 4 r_1r_2 \sin^2 \psi \\ 4c^2 & = (2a)^2 - 4 r_1 r_2 \sin^2 \psi \\ 4r_1 r_2 \sin^2 \psi &= 4a^2 -4c^2 = 4b^2 \\ r_1 r_2 &= \frac{b^2}{\sin^2 \psi} \end{align*}

Edit: If you wished to handle the ellipse and hyperbola simultaneously it would perhaps be best to call the angle in the triangle opposite the major axis $\theta$, so that for the ellipse $\theta = \pi-2\psi$ and for the hyperbola $\theta = 2\psi$. \begin{align*} (2c)^2 &= r_1^2 + r_2^2 - 2r_1r_2 \cos \theta \\ 4c^2 &= r_1^2 +r_2^2 \pm 2 r_1 r_2 \cos 2\psi \quad \text{(taking + for the ellipse and - for the hyperbola)}\\ 4c^2 &= r_1^2 + r_2 ^2 \pm 2r_1r_2(1-2 \sin ^2 \psi) \\ 4c^2 &= (r_1\pm r_2)^2 \mp 4 r_1r_2 \sin^2 \psi \\ 4c^2 & = (2a)^2 \mp 4 r_1 r_2 \sin^2 \psi \\ \pm4r_1 r_2 \sin^2 \psi &= 4a^2 -4c^2 = \pm 4b^2 \\ r_1 r_2 &= \frac{b^2}{\sin^2 \psi} \end{align*}