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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