# If p and q are the lengths of perpendiculars from the origin to the lines x cos θ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2

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 Comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain

A = cos θ, B = -sin θ, and C = -k cos 2θ

Given that p is the length of the perpendicular from (0, 0) to line (1). p = k cos 2θ

Squaring both sides

P2 = k2 cos22θ …………………(3)

Comparing equation (2) to the general equation of line i.e., Ax + By + C = 0, we obtain

A = sec θ, B = cosec θ, and C = -k

Given that q is the length of the perpendicular from (0, 0) to line (2)   Multiplying both sides by 2

2q = 2k cos θ sin θ = k × 2sinθ cosθ

2q = k sin 2θ

Squaring both sides

4q2 = k2 sin22θ …………………(4)

Adding (3) and (4) we get

p2 + 4q2 = k2 cos2 2θ + k2 sin2

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

p2 + 4q2 = k2

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