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.

Check out my analysis of your question.
Equation is 2xcosα−3ysinα=0. This equation passes through origin and points A and B are on x axis. Assume slope of equation be tan $\phi$.

I have illustrated the diagram here

now product of length of two perpendiculars will be 5 sin$^2$ $\phi$. From equation of line slope of line is $$tan\phi =\frac{2cos\alpha}{3sin\alpha}$$

get value of $sin^2\phi$ from $tan\phi$. You will see it is not free of alpha.You can see that in graph too here  (I have updated it). As graph is symmetrical ,length will depend on $\phi$.
Maybe I interpreted the question incorrectly. Correct me if it's wrong
