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Let $\mathrm{P}(3 \sec \theta, 2 \tan \theta) \mathrm{Q}(3 \sec \phi, 2 \tan \phi)$ be two points on $\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1$ such that $\theta+\phi=\frac{\pi}{2} .0 < \theta, \phi < \frac{\pi}{2}$. Then the ordinate of the point of intersection of the normals at $P$ and $Q$ is
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The correct answer is:
$-\frac{13}{2}$
$\begin{aligned} & \frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2, P(3 \sec \theta, 2 \tan \theta) \\ \therefore & \frac{9 x}{3 \sec \theta}+\frac{4 y}{2 \tan \theta}=9+4=13 \\ \Rightarrow & 3 x \cos \theta+2 y \cot \theta=13 \end{aligned}$
Similarly, $3 x \cos \phi+2 y \cot \phi=13$
$3 x \cos \left(\frac{\pi}{2}-\theta\right)+2 y \cot \left(\frac{\pi}{2}-\theta\right)=13$
$\Rightarrow 3 x \sin \theta+2 y \tan \theta=13 \ldots \ldots(2)$
$\therefore \frac{13-2 \mathrm{y} \cot \theta}{3 \cos \theta}=\frac{13-2 \mathrm{y} \tan \theta}{3 \sin \theta}$
$\Rightarrow 13 \sin \theta-2 y \cos \theta=13 \cos \theta-2 y \sin \theta$
$\Rightarrow 13 \tan \theta-2 y=13-2 y \tan \theta$
$\Rightarrow 13(\tan \theta-1)=-2 y(\tan \theta-1)$
$\therefore \mathrm{y}=-\frac{13}{2}$
$\therefore$ Ordinate $=-\frac{13}{2}$
Similarly, $3 x \cos \phi+2 y \cot \phi=13$
$3 x \cos \left(\frac{\pi}{2}-\theta\right)+2 y \cot \left(\frac{\pi}{2}-\theta\right)=13$
$\Rightarrow 3 x \sin \theta+2 y \tan \theta=13 \ldots \ldots(2)$
$\therefore \frac{13-2 \mathrm{y} \cot \theta}{3 \cos \theta}=\frac{13-2 \mathrm{y} \tan \theta}{3 \sin \theta}$
$\Rightarrow 13 \sin \theta-2 y \cos \theta=13 \cos \theta-2 y \sin \theta$
$\Rightarrow 13 \tan \theta-2 y=13-2 y \tan \theta$
$\Rightarrow 13(\tan \theta-1)=-2 y(\tan \theta-1)$
$\therefore \mathrm{y}=-\frac{13}{2}$
$\therefore$ Ordinate $=-\frac{13}{2}$
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