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Product CF*PG for Tangent Perpendicular and Normal | JEE

JEE Maths question with a full step-by-step solution.

Question
If CFCF is the perpendicular from the centre CC of the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 on the tangent at a point PP, and GG is the point where the normal at PP meets the major axis, then CF⋅PGCF\cdot PG is:
Ab2b^2correct
B2b22b^2
Cb22\dfrac{b^2}{2}
D3b23b^2
Solution
Step 1: Tangent at P=(acos⁡θ,bsin⁡θ)P=(a\cos\theta,b\sin\theta) is xacos⁡θ+ybsin⁡θ=1\dfrac{x}{a}\cos\theta+\dfrac{y}{b}\sin\theta=1. The perpendicular from the centre is
CF=aba2sin⁡2θ+b2cos⁡2θ.CF=\frac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}.
 Step 2: The normal at PP meets the major axis at GG; the length from PP to GG is
PG=baa2sin⁡2θ+b2cos⁡2θ.PG=\frac{b}{a}\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}.
 Step 3: Multiply.
CF⋅PG=aba2sin⁡2θ+b2cos⁡2θ⋅baa2sin⁡2θ+b2cos⁡2θ=b2.CF\cdot PG=\frac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}\cdot\frac{b}{a}\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}=b^2.
 Correct answer: (1)
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