If p and q are the lengths of the perpendiculars from the origin on the lines,
x cosec α - y sec α = k cot 2α and
x sinα + y cosα = k sin2α
respectively, then k² is equal to :

Solution

Maths sir · 3 steps

Clear the cosec and sec first, and both lines take the neat form x cos α ± y sin α.

1. Tidy the first line. Multiply x cosec α - y sec α = k cot 2α by sin α cos α. The left side becomes x cos α - y sin α. The right side becomes k(cos 2α)/(sin 2α) × (sin 2α)/2 = k/2 cos 2α.
2. Find p and q. The distance from the origin to ax + by = c is |c|/√(a²+b²). Both lines have a² + b² = cos² α + sin² α = 1, so p = (|k cos 2α|)/2 and q = |k sin 2α|.

   (2p)² + q² = k²cos² 2α + k²sin² 2α = k²

3. So. k² = 4p² + q², which is option (a).

The idea to keep. When one distance carries cos 2α and the other sin 2α, match the coefficients, square and add. cos² θ + sin² θ = 1 removes α.

Answer from the NTA final answer key. Solution by PrepWiser.