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Eccentricity of Locus of Midpoint of Normal Intercepts | JEE

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

Question

The normal at a variable point

P

on an ellipse

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

of eccentricity

e

meets the axes in

Q

and

R

. Then the locus of the midpoint of

QR

is a conic with eccentricity

e'

such that:

e'

is independent of

e

e'=1

e'=e

correct

e'=\dfrac{1}{e}

Solution

Step 1: Normal at

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

\dfrac{a^2x}{a\cos\theta}-\dfrac{b^2y}{b\sin\theta}=a^2-b^2

, i.e.

ax\sec\theta-by\csc\theta=a^2-b^2

. Step 2: Intercepts on the axes:

Q=\left(\frac{(a^2-b^2)\cos\theta}{a},\,0\right),\qquad R=\left(0,\,-\frac{(a^2-b^2)\sin\theta}{b}\right).

Step 3: Let the midpoint be

(h,k)

h=\frac{(a^2-b^2)\cos\theta}{2a},\qquad k=-\frac{(a^2-b^2)\sin\theta}{2b}.

Eliminating

\theta

\frac{a^2x^2}{\left(\tfrac{a^2-b^2}{2}\right)^2}+\frac{b^2y^2}{\left(\tfrac{a^2-b^2}{2}\right)^2}=1.

Step 4: This ellipse has semi-axes

A=\dfrac{a^2-b^2}{2a},\ B=\dfrac{a^2-b^2}{2b}

with

B>A

. Its eccentricity satisfies

e'^2=1-\dfrac{A^2}{B^2}=1-\dfrac{b^2}{a^2}=e^2

, so

e'=e

. Correct answer: (3)

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