Solving a problem in Coordinate Geometry Suppose,  $2x \cos \alpha - 3y \sin \alpha = 6$ is equation of a variable straight line. From two point $A(\sqrt{5},0$) and  $B(-\sqrt{5},0)$ foot of the altitude on that straight line is $P$ and $Q$ respectively. Show that the product of the length of two line segment $AP$ and $BQ$ is free of $\alpha$.
This question appeared in my exam today. The way I did it is first constructed the equation of two perpendicular line to $2x \cos \alpha - 3y \sin \alpha = 6$ which goes through the points $A(\sqrt{5},0$) and  $B(-\sqrt{5},0)$. In this way, for $A$ and $B$, I got the following equations respectively-
$$3x \sin \alpha + 2y \cos \alpha -3 \sqrt{5}\sin \alpha=0 \qquad(1)$$$$3x \sin \alpha + 2y \cos \alpha +3 \sqrt{5}\sin \alpha=0 \qquad(2)$$
Then I found that the line $(1)$ intersects the line $2x \cos \alpha - 3y \sin \alpha = 6$ at point $$P\left ( \frac{9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}, \frac{-18 \sin \alpha+6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )$$
and the line $(2)$ intersects the line $2x \cos \alpha - 3y \sin \alpha = 6$ at point $$Q\left ( \frac{-9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}, \frac{-18 \sin \alpha-6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )$$
Now using distance formula $$AP = \sqrt{ \left ( \frac{9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}- \sqrt5 \right )^2 + \left ( \frac{-18 \sin \alpha+6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )^2}$$ and $$ BQ= \sqrt{ \left ( \frac{-9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} +\sqrt5 \right )^2 + \left ( \frac{-18 \sin \alpha-6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )^2}$$
Now the multiplication product $AP \cdot BQ$ indeed gives a constant value of $4$ which is free of the arbitrary variable $\alpha$ as you can see here is the simplified version of product of those two quantity. But this is tedious and I do not think this is the only way to do it and an appropriate way to follow in exam with limited time. So, I am looking for an alternative, time saving proof of it. I was wondering if using parametric form would help me, but I think it would get as difficult equally.
I created a visualisation for you on desmos to help my problem understand better.

$ \displaystyle 2x \cos \alpha - 3y \sin \alpha = 6 \implies y = \frac{2 \cot \alpha}{3} x - 2 \csc \alpha = mx + c$ where $m = \tan \theta = \dfrac{2 \cot \alpha}{3}$
Consider $0 \leq \alpha \leq 2\pi$ that gives all possible lines for the given equation. If $\alpha = \frac{\pi}{2}$ or $\frac{3 \pi}{2}$, the line is $y = -2$ or $y = 2$ respectively and the perp distance from both points on x-axis would be $2$. So the product is $4$. For other values of $\alpha$
The line intersects x-axis at $~ \displaystyle x = \frac{3}{\cos\alpha}$
So distance of the given points to the intersection point is,
$ \left|\displaystyle \sqrt5 \pm \frac{3}{\cos\alpha}\right|$
The product of perpendicular lengths to the line is then given by,
$ \displaystyle \left| 5 - \frac{9}{\cos^2\alpha}\right| \cdot \sin^2\theta = \left| \frac{5 \cos^2\alpha - 9}{\cos^2\alpha} \cdot \frac{\tan^2\theta}{1 + \tan^2\theta} \right|$
Now, $ \displaystyle \frac{\tan^2\theta}{1 + \tan^2\theta} = \frac{4 \cot^2\alpha}{9 + 4 \cot^2\alpha} = \frac{4 \cos^2\alpha}{9 \sin^2\alpha + 4 \cos^2\alpha}$
$ \displaystyle = \frac{4 \cos^2\alpha}{9 - 5 \cos^2\alpha}$
That leads to product being $4$.
