Animated Solution for Mathematics - Straight Lines: If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines, $x\mathrm{cosec}\alpha -y\sec \alpha =k\cot 2\alpha$ and $x\sin \alpha +y\cos \alpha =k\sin 2\alpha$ respectively, then $k^2$ is equal to:

Taming the Trigonometric Beast

Welcome, fellow traveler on the JEE journey. Today, we are going to tackle a problem that, at first glance, looks like a chaotic mess of trigonometric functions.

We have two lines, $L_1$ and $L_2$, defined by complex-looking equations involving $\csc \alpha$, $\sec \alpha$, and $\cot 2\alpha$. But here is the secret: in the world of JEE Advanced, complexity is often just a mask for underlying simplicity. Our mission is to peel back that mask.

We are given that $p$ and $q$ are the perpendicular distances from the origin to these lines. Whenever you see "perpendicular distance from the origin," your mind should immediately jump to the standard distance formula.

For any line $Ax+By+C=0$, the distance $d$ from the origin $(0,0)$ is simply:

This is our North Star. Our entire strategy is to manipulate the given equations into this standard $Ax+By+C=0$ form. Let's start with the first line, $L_1$: $x\csc \alpha -y\sec \alpha =k\cot 2\alpha$.

This equation looks intimidating, doesn't it? Let's break it down by converting the trigonometric terms into their sine and cosine counterparts:

Now, let's find a common denominator on the left-hand side. By cross-multiplying, we get:

Here is where the magic happens. Remember the double-angle identity: $\sin 2\alpha =2\sin \alpha \cos \alpha$. If we substitute this into the denominator on the right, the $\sin \alpha \cos \alpha$ terms on both sides cancel out perfectly!

We are left with:

Bringing everything to one side, we get the standard form: $x\cos \alpha -y\sin \alpha -\frac{k}{2}\cos 2\alpha =0$. Now, applying our distance formula for $p$:

Since ${\cos }^2\alpha +{\sin }^2\alpha =1$, the denominator becomes $\sqrt{1}=1$. Thus, $p=|\frac{k}{2}\cos 2\alpha |$, or $2p=|k\cos 2\alpha |$.

Now for $L_2$: $x\sin \alpha +y\cos \alpha =k\sin 2\alpha$. This one is already much friendlier. In standard form, it is $x\sin \alpha +y\cos \alpha -k\sin 2\alpha =0$.

Applying the distance formula for $q$:

Again, the denominator is $1$. So, $q=|k\sin 2\alpha |$.

We have arrived at our two simplified equations:

1) $2p=|k\cos 2\alpha |$

2) $q=|k\sin 2\alpha |$

We need to find $k^2$ and eliminate $\alpha$. When you have $\sin$ and $\cos$ of the same angle, squaring and adding is the ultimate weapon. Let's square both equations:

Adding them together gives us:

And there it is—the final, beautiful cancellation. Since ${\cos }^{2}2\alpha +{\sin }^{2}2\alpha =1$, we are left with $k^2=4p^2+q^2$.

We have successfully navigated the trigonometric maze and arrived at the solution. Remember, in physics and math, the most complex problems often yield to the simplest, most fundamental principles. Keep practicing, and keep looking for that elegance!