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Product CF*PG for Tangent Perpendicular and Normal | JEE

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

Question

CF

is the perpendicular from the centre

C

of the ellipse

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

on the tangent at a point

P

, and

G

is the point where the normal at

P

meets the major axis, then

CF\cdot PG

is:

b^2

correct

2b^2

\dfrac{b^2}{2}

3b^2

Solution

Step 1: Tangent at

P=(a\cos\theta,b\sin\theta)

\dfrac{x}{a}\cos\theta+\dfrac{y}{b}\sin\theta=1

. The perpendicular from the centre is

CF=\frac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}.

Step 2: The normal at

P

meets the major axis at

G

; the length from

P

G

PG=\frac{b}{a}\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}.

Step 3: Multiply.

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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