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.
 A: Well I have another solution which is very helpful for exams based on mcq pattern, like the one we give it in india.
I have learnt and derived a property in conic sections which says that in a ellipse, the product of perpendiculars from both foci of ellipse to tangent is equal to b$^2$ ,b is semi minor axis.
equation of a general ellipse is

$\frac{x^2}{a^2}$ + $\frac{y^2}{b^2}$=1

it's tangent equation is-

$\frac{xx1}{a^2}$ + $\frac{yy1}{b^2}$=1 (x1 and y1 are points on ellipse, which is taken x1=acos$\phi$ and y1=bsin$\phi$)

now tangents is revolving will all possible slopes around ellipse...now to ease our work assume slope =o for a tangent... $\alpha$ in your equation will be $\pi$/2 for zero slope with y intercept +2 or-2
.put that in your equation and find the product of perpendiculars, as line is horizontal ,perpendicular will be of  of equal length. you will find it to be 4.
now as I told product of perpendiculars is $b^2$, so b=2 ,now use that general tangent equation and given equation to find value of a=3.you will see that it satisfies the equation and a ellipse. It means we were assuming it right. The parameter in your equation was making the arbitrary line a tangent to ellipse.
Here is illustration for better understanding
image
A: $ \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 $4$.
A: 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 $B$.
A: 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
