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← JEE Main 2014 · 11 Apr · Shift 1
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JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 82

Let P(3sec⁡θ,2tan⁡θ)P(3\sec\theta,2\tan\theta) and Q(3sec⁡ϕ,2tan⁡ϕ)Q(3\sec\phi,2\tan\phi) where θ+ϕ=π2\theta+\phi=\dfrac\pi2, be two distinct points on the hyperbola x29−y24=1\dfrac{x^2}9-\dfrac{y^2}4=1. Then the ordinate of the point of intersection of the normals at P and Q is :

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Answer and worked solution

Official answer: D

The normal at (3sec⁡θ,2tan⁡θ)(3\sec\theta,2\tan\theta) is 3xcos⁡θ+2ycot⁡θ=133x\cos\theta+2y\cot\theta=13. At ϕ=π2−θ\phi=\frac\pi2-\theta it is 3xsin⁡θ+2ytan⁡θ=133x\sin\theta+2y\tan\theta=13.

Eliminating xx: cot⁡θ (13−2ytan⁡θ)=13−2ycot⁡θ\cot\theta\,(13-2y\tan\theta)=13-2y\cot\theta, so 13(cot⁡θ−1)=−2y(cot⁡θ−1)13(\cot\theta-1)=-2y(\cot\theta-1). With θ≠π4\theta\ne\frac\pi4 (the points are distinct), y=−132y=-\dfrac{13}2.

Topics:Tangents and normals to conicsHyperbola
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