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.

One missing factor is the locus of points $P$ and $Q$.
For point P we have the system of equations
$2\cos\alpha x - 3\sin\alpha y=6$
$3\sin\alpha x + 2\cos\alpha y=3\sqrt5\sin\alpha$
Square both equations and add. Cross-product terms cancel and we end with
$(4\cos^2\alpha+9\sin^2\alpha)(x^2+y^2)=36+45\sin^2\alpha$
Eliminate $\cos\alpha$ using $\cos^2\alpha+\sin^2\alpha=1$, then
$(4+5\sin^2\alpha)(x^2+y^2)=36+45\sin^2\alpha$
$\color{blue}{x^2+y^2=9, \text{ a circle of radius 3}}$
Similarly $Q$ has this same locus.
We then draw this circle which intersects the $x$-axis at $C=(3,0)$ and $D=(-3,0)$. With this in hand we extend $\overline{AP}$ through $A$ to point $Q'$. By symmetry $AQ'=BQ$ (the lines $\overline{AP},\overline{BQ}$ are parallel and equidistant from the center, so $Q$ and $Q'$ are antipodes on the circle). At the same time for intersecting chords of a circle $(AP)(AQ')=(AC)(AD)$. So:
$(AP)(BQ)=(AP)(AQ')=(AC)(AD)=(3-\sqrt5)(3+\sqrt5)=4.$

This exercise illustrates a  theorem from the geometry of conic sections. Given an ellipse or a hyperbola, we can drop a perpendicular from both foci to any tangent of the conic. The product of the altitudes will then be the square of the semiminor axis for an ellipse or the square of the semiconjugate axis for a hyperbola. We may interpret the latter case as presenting a negative product (the square of a pure imaginary semiminor axis) because the altitudes are oppositely oriented as if one of them were negative. The family of lines in the problem at hand are the tangents to an ellipse whose vertices are $(\pm3,0)$ (matching a diameter of the circular locus) with foci at the same points as those chosen for $A$ and $B$.
