If p and q are the lengths of perpendiculars from the origin to the lines x cosθ − y sinθ = k cos2θ and x secθ + y cscθ = k, prove that p² + 4q² = k².

x cos θ - y sin θ = k cos 2θ   ... (1)

x sec θ + y csc θ = k   ... (2)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁,y₁) is given by d = ^{|Ax₁ + By₁ + C|}/_{√(A² + B²)}

On comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain A = cosθ, B = -sinθ, and C = -kcos2θ.

It is given that p is the length of the perpendicular from (0, 0) to line ………..(1)

A = cos θ, B = -sin θ, C = -k cos 2θ p₁ = ^{|-k cos 2θ|}/_{√(cos²θ + sin²θ)} = |k cos 2θ|   ... (3)

A = sec θ, B = csc θ, C = -k

p₂ = ^|-k|/_{√(sec²θ + csc²θ)} = ^|k|/_{√(sec²θ + csc²θ)}   ... (4)

p₁² + 4p₂² = (|k cos 2θ|)² + 4(^|k|/_{√(sec²θ + csc²θ)})²

= k² cos² 2θ + ^4k2/_{sec²θ + csc²θ}

= k² cos² 2θ + ^4k2/_{¹/cos²θ + ¹/sin²θ}

= k² cos² 2θ + ^{4k2 sin2θ cos2θ}/_{sin²θ + cos²θ}

= k² cos² 2θ + 4k² sin²θ cos²θ

= k² cos² 2θ + k²(2 sin θ cos θ)²

= k² cos² 2θ + k² sin² 2θ

= k²(cos² 2θ + sin² 2θ)

= k²