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Eccentricity from Tangent Chord on Auxiliary Circle | JEE

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

Question

If the tangent at a point

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

on the ellipse

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

meets the auxiliary circle in two points and the chord joining them subtends a right angle at the centre, then the eccentricity is:

(1+\cos^2\theta)^{-1/2}

1+\sin^2\theta

(1+\sin^2\theta)^{-1/2}

correct

1+\cos^2\theta

Solution

Step 1: The tangent is

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

. Homogenise the auxiliary circle

x^2+y^2=a^2

with this line:

x^2+y^2=a^2\left(\frac{x}{a}\cos\theta+\frac{y}{b}\sin\theta\right)^2.

This gives the pair of lines joining the centre to the two intersection points. Step 2: For these lines to be perpendicular, the sum of the coefficients of

x^2

and

y^2

must vanish.

(1-\cos^2\theta)+\left(1-\frac{a^2\sin^2\theta}{b^2}\right)=0 \;\Rightarrow\; 1+\sin^2\theta=\frac{a^2\sin^2\theta}{b^2}.

Step 3: Hence

\dfrac{b^2}{a^2}=\dfrac{\sin^2\theta}{1+\sin^2\theta}

, so

e^2=1-\frac{b^2}{a^2}=1-\frac{\sin^2\theta}{1+\sin^2\theta}=\frac{1}{1+\sin^2\theta}\;\Rightarrow\; e=(1+\sin^2\theta)^{-1/2}.

Correct answer: (3)

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