Given:

The equations of given lines are

x cos θ – y sin θ = k cos 2θ …………………… (1)

x sec θ + y cosec θ = k ……………….… (2)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

q = k cos θ sin θ

Multiply both sides by 2, we get

2q = 2k cos θ sin θ = k × 2sin θ cos θ

2q = k sin 2θ

Squaring both sides, we get

4q² = k² sin²2θ …………………(4)

Now add (3) and (4) we get

p² + 4q² = k² cos² 2θ + k² sin² 2θ

p² + 4q² = k² (cos² 2θ + sin² 2θ) [Since, cos² 2θ + sin² 2θ = 1]

∴ p² + 4q² = k²

Hence proved.