Conics - required to show $SR \times S'R' = b^2$

Question

Consider the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b > 0$.

$R$ and $R'$ are the feet of the perpendiculars from the foci $S$ and $S'$ on to the tangent at $P(a\cos\theta, b\sin\theta)$.

Show that $SR\times S'R'=b^2$.

(You may assume the tangent to ellipse at the point $P(a\cos\theta,b\sin\theta)$ has equation $bx\cos\theta+ay\sin\theta-ab=0$.)

so basically in this question, i know you have to do the perpendicular distances of SR and S'R' but then after that, I'm not sure how to go about simplifying it to get the answer. thanks for the help :)

Answer 1

We have $bx\cos\theta+ay\sin\theta-ab=0$. Distances from $(\pm ae,0)$ would be: $$\frac{|b(\pm ae)\cos\theta+a(0)\sin\theta-ab|}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}}=\frac{ab(1\mp e\cos\theta)}{\sqrt{a^2\sin^2\theta+a^2(1-e^2)\cos^2\theta}}=\frac{ab(1\mp e\cos\theta)}{a\sqrt{1-e^2\cos^2\theta}}$$ Product of which is: $$\frac{a^2b^2(1-e^2\cos^2\theta)}{a^2(1-e^2\cos^2\theta)}=b^2$$

Answer 2

The perpendicular distances of the tangent from the foci$(\pm ae,0)$

$$\frac{b\cos\theta(\pm ae)-ab}{\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}}$$

So, the product $$=\frac{a^2b^2(1-e^2\cos^2\theta)}{a^2\cos^2\theta+b^2\sin^2\theta}$$

Use $b^2=a^2(1-e^2)\iff e^2=\cdots$ and $\sin^2\theta=1-\cos^2\theta$