Q.If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x\cos\theta - y\sin\theta = k\cos 2\theta$ and $x\sec\theta + y\,\text{cosec}\,\theta = k$, respectively, prove that $p^2 + 4q^2 = k^2$.

Concept: Area of Triangle from Lines — here, the perpendicular distance formula is the key.

Step 1: For the first line $x\cos\theta - y\sin\theta = k\cos 2\theta$, rewrite in normal form:

Coefficient of $x$ is $\cos\theta$, of $y$ is $-\sin\theta$, so $\sqrt{\cos^2\theta + \sin^2\theta} = 1$.

Thus the perpendicular distance from the origin is

$p = \frac{|k\cos 2\theta|}{1} = |k\cos 2\theta|$.

Step 2: For the second line $x\sec\theta + y\,\text{cosec}\,\theta = k$, rewrite as

$\frac{x}{\cos\theta} + \frac{y}{\sin\theta} = k$.

Multiply through by $\sin\theta\cos\theta$:

$x\sin\theta + y\cos\theta = k\sin\theta\cos\theta$.

Step 3: Normalise: $\sqrt{\sin^2\theta + \cos^2\theta} = 1$, so the perpendicular distance from the origin is …

Concept and Intuition

When a problem asks for the perpendicular distance from the origin to a line, the standard formula is your friend: for a line $ax + by + c = 0$, the distance from $(0,0)$ is $\frac{|c|}{\sqrt{a^2 + b^2}}$. Here, both lines are given in forms that look different — one has $\cos\theta$ and $\sin\theta$, the other has $\sec\theta$ and $\csc\theta$. The trick is to rewrite each line in the standard form, compute $p$ and $q$, then combine them.

The result $p^2 + 4q^2 = k^2$ is neat because it's independent of $\theta$ — the trigonometric terms cancel out completely. That's the sign of a well-constructed identity.

Step-by-Step Solution

1. First line: $x\cos\theta - y\sin\theta = k\cos 2\theta$

Rewrite in standard form $ax + by + c = 0$:

Here $a = \cos\theta$, $b = -\sin\theta$, $c = -k\cos 2\theta$.

The perpendicular distance $p$ from the origin is:

Since $\cos^2\theta + \sin^2\theta = 1$, the denominator is $1$. Also, $k$ is presumably positive (length), so:

2. Second line: $x\sec\theta + y\,\text{cosec}\,\theta = k$

Rewrite in standard form:

Here $a = \sec\theta$, $b = \text{cosec}\,\theta$, $c = -k$.

The perpendicular distance $q$ from the origin is:

So $q = \frac{k}{\sqrt{\sec^2\theta + \text{cosec}^2\theta}}$.

3. Simplify the denominator for $q$

Recall $\sec^2\theta = \frac{1}{\cos^2\theta}$ and $\text{cosec}^2\theta = \frac{1}{\sin^2\theta}$. So:

Therefore:

Thus:

Again, squaring removes the absolute value:

4. Compute $p^2 + 4q^2$

We have:

$p^2 = k^2 \cos^2 2\theta$ …