Question -
Answer -
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.