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Eccentricity from Tangent Chord on Auxiliary Circle | JEE

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Question
If the tangent at a point (acos⁡θ,bsin⁡θ)(a\cos\theta,b\sin\theta) on the ellipse x2a2+y2b2=1\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:
A(1+cos⁡2θ)−1/2(1+\cos^2\theta)^{-1/2}
B1+sin⁡2θ1+\sin^2\theta
C(1+sin⁡2θ)−1/2(1+\sin^2\theta)^{-1/2}correct
D1+cos⁡2θ1+\cos^2\theta
Solution
Step 1: The tangent is xacos⁡θ+ybsin⁡θ=1\dfrac{x}{a}\cos\theta+\dfrac{y}{b}\sin\theta=1. Homogenise the auxiliary circle x2+y2=a2x^2+y^2=a^2 with this line:
x2+y2=a2(xacos⁡θ+ybsin⁡θ)2.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 x2x^2 and y2y^2 must vanish.
(1−cos⁡2θ)+(1−a2sin⁡2θb2)=0  ⇒  1+sin⁡2θ=a2sin⁡2θb2.(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 b2a2=sin⁡2θ1+sin⁡2θ\dfrac{b^2}{a^2}=\dfrac{\sin^2\theta}{1+\sin^2\theta}, so
e2=1−b2a2=1−sin⁡2θ1+sin⁡2θ=11+sin⁡2θ  ⇒  e=(1+sin⁡2θ)−1/2.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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