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Mathematics Question on Conic sections

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

A

113\frac{11}{3}

B

−113\frac{-11}{3}

C

132\frac{13}{2}

D

−132\frac{-13}{2}

Answer

−132\frac{-13}{2}

Explanation

Solution

p(3 sec θ, 2 tan θ)Q=(3 sec ϕ , 2 tanϕ)p \left(3 \,sec\, \theta,\,2 \,tan \,\theta\right) \quad Q = \left(3 \,sec\,\phi\, ,\, 2 \,tan\phi\right)
θ+ϕ=π2Q=(3 cosec θ, 2 cotθ)\theta +\phi = \frac{\pi}{2}\quad Q = \left(3 \,cosec \, \theta , \,2 \, cot\theta\right)
Equation of normal at p=p =
=3x cos θ+2y cot θ=13= 3x \, cos \,\theta + 2y \, cot \,\theta=13
=3x sin θ cos θ+2y cos θ=13 sinθ...(1)= 3x \, sin \,\theta \, cos \,\theta + 2y \,cos \,\theta = 13 \,sin\theta\quad ...\left(1\right)
equation of normal at Q ⇒\Rightarrow
=3x sin θ+2y tan θ=13= 3x\, sin \, \theta + 2y \, tan \, \theta = 13
=3x sin θ cos θ+2y sin θ=13 cosθ...(2)= 3x\, sin\, \theta \, cos \, \theta + 2y \, sin \, \theta = 13\, cos\theta \quad ...\left(2\right)
(1)−(2) ⇒\left(1\right)-\left(2\right)\, \Rightarrow
2y (cos θ−sin θ)=13(sin θ−cos θ)2y \, \left(cos\, \theta - sin \, \theta \right) = 13 \left(sin\, \theta - cos \,\theta \right)
2y=−13⇒−1322y = -13 \Rightarrow \frac{-13}{2}