On the plane we have two points $A(\sqrt{a^2-b^2},0),B(-\sqrt{a^2-b^2},0)$ with $a>b>0$ and the line $L$, of which the equation is given as:$$\dfrac{x\cos{\theta}}{a}+\dfrac{y\sin{\theta}}{b}=1$$
Define $d_{1}$ to be the distance between the straight line $L$ and point $A$ and $d_{2}$ to be the distance between the straight line $L$ and point $B$. Find the value $$d_{1}\cdot d_{2}$$
My attempt:
Since
$$d_{1}=\dfrac{\dfrac{\sqrt{a^2-b^2}\cos{\theta}}{a}-1}{\sqrt{\dfrac{\cos^2{\theta}}{a^2}+\dfrac{\sin^2{\theta}}{b^2}}}$$
and
$$d_{2}=\dfrac{\dfrac{-\sqrt{a^2-b^2}\cos{\theta}}{a}-1}{\sqrt{\dfrac{\cos^2{\theta}}{a^2}+\dfrac{\sin^2{\theta}}{b^2}}}$$
therefore
$$d_{1}d_{2}=\dfrac{1-\dfrac{a^2-b^2}{a^2}\cos^2{\theta}}{\dfrac{\cos^2{\theta}}{a^2}+\dfrac{\sin^2{\theta}}{b^2}}=\frac{\sin^2{\theta}+\dfrac{b^2}{a^2}\cos^2{\theta}}{\dfrac{b^2\cos^2{\theta}+a^2\sin^2{\theta}}{a^2b^2}}=b^2$$
I think this is very ugly. Can someone do this via another method? Thank you.
The basic process won't get much better than you have, but the presentation can be made a little cleaner with a couple of minor adjustments to avoid having to write so many radicals and fractions:
Define $c := \sqrt{a^2-b^2}$, so that $A = (c,0)$ and $B=(-c,0)$, and write the equation of $L$ as $$x b \cos\theta + y a \sin\theta = a b$$ so that $$\begin{align} d_1 &= \frac{b\;|c\cos\theta-a|}{\sqrt{a^2\sin^2\theta + b^2\cos^2\theta}} = \frac{b\;\left( a - c\cos\theta\right)}{\sqrt{a^2\sin^2\theta + b^2\cos^2\theta}} \quad \text{since}\quad c\cos\theta \leq c \leq a\\[6pt] d_2 &= \frac{b\;\left( a + c\cos\theta\right)}{\sqrt{a^2\sin^2\theta + b^2\cos^2\theta}} \end{align}$$ Then $$d_1 d_2 = \frac{b^2\;\left( a^2 - c^2\cos^2\theta \right)}{a^2\sin^2\theta + b^2\cos^2\theta}$$
Now, observe that $$a^2 - c^2\cos^2\theta \;\;=\;\; a^2 - (a^2-b^2)\cos^2\theta \;\;=\;\; a^2 \sin^2\theta + b^2\cos^2\theta$$ so that $$d_1 d_2 = b^2$$