We need to prove that $p^2 + 4q^2 = k^2$.

Step 1: Find $p$ for the first line

The first line is $x \cos \theta - y \sin \theta = k \cos 2\theta$, which can be written as:

$$x \cos \theta - y \sin \theta - k \cos 2\theta = 0$$

The perpendicular distance from origin $(0, 0)$ to line $ax + by + c = 0$ is given by:

$$p = \frac{|c|}{\sqrt{a^2 + b^2}}$$

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

Therefore:

$$p = \frac{|-k \cos 2\theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = \frac{|k \cos 2\theta|}{\sqrt{1}} = |k \cos 2\theta|$$

Squaring both sides:

$$p^2 = k^2 \cos^2 2\theta$$

Step 2: Find $q$ for the second line

The second line is $x \sec \theta + y \cosec \theta = k$, which can be written as:

$$x \sec \theta + y \cosec \theta - k = 0$$

Here, $a = \sec \theta$, $b = \cosec \theta$, $c = -k$

Therefore:

$$q = \frac{|-k|}{\sqrt{\sec^2 \theta + \cosec^2 \theta}} = \frac{|k|}{\sqrt{\sec^2 \theta + \cosec^2 \theta}}$$

Now,

$$\sec^2 \theta + \cosec^2 \theta = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} = \frac{1}{\cos^2 \theta \sin^2 \theta}$$

So,

$$\sqrt{\sec^2 \theta + \cosec^2 \theta} = \frac{1}{|\cos \theta \sin \theta|} = \frac{2}{|\sin 2\theta|}$$

Therefore:

$$q = \frac{|k| \cdot |\sin 2\theta|}{2}$$

Squaring both sides:

$$q^2 = \frac{k^2 \sin^2 2\theta}{4}$$

Step 3: Prove $p^2 + 4q^2 = k^2$

Now,

$$p^2 + 4q^2 = k^2 \cos^2 2\theta + 4 \times \frac{k^2 \sin^2 2\theta}{4}$$ $$= k^2 \cos^2 2\theta + k^2 \sin^2 2\theta$$ $$= k^2 (\cos^2 2\theta + \sin^2 2\theta)$$ $$= k^2 \times 1 = k^2$$

∴ Hence proved.