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Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ be two points such that $\theta+\phi=\frac{\pi}{2}$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$, then $k=$
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The correct answer is:
$-\left(\frac{a^2+b^2}{b}\right)$
The equation of the hyperbola is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The equation of the normal at a point $P(a \sec \theta, b \tan \theta)$ is given $\frac{b y}{\tan \theta}+\frac{a x}{\sec \theta}=a^2+b^2$ $\Rightarrow \quad b \cot \theta y+a \cos \theta x-\left(a^2+b^2\right)=0$
The equation of normal at a point $\theta(a \sec \phi, b \tan \phi)$ is given
$\begin{array}{r}\frac{b y}{\tan \phi}+\frac{a x}{\sec \phi}=a^2+b^2 \\ \Rightarrow b \cot \phi y+a \cos \phi x-\left(a^2+b^2\right)=0\end{array}$
Solving Eqs. (i) and (ii), we get
$\begin{aligned} & \frac{x}{-\left(a^2+b^2\right) b \cot \theta+\left(a^2+b^2\right) b \cot \phi} \\ & =\frac{-y}{-\left(a^2+b^2\right) a \cos \phi+\left(a^2+b^2\right) a \cos \theta} \\ & =\frac{1}{b \cot \theta a \cos \phi-b \cot \phi a \cos \theta} \\ & \Rightarrow \frac{-y}{-\left(a^2+b^2\right) a \cos \phi+\left(a^2+b^2\right) a \cos \theta} \\ & =\frac{1}{b \cot \theta a \cos \phi-b \cot \phi a \cos \theta} \\ & \Rightarrow \frac{-y}{a\left(a^2+b^2\right)[\cos \theta-\cos \phi} \\ & =\frac{1}{a b[\cot \theta(a) \cos \phi-\cos \phi(a) \cos \theta} \\ & \Rightarrow y=\frac{-\left(a^2+b^2\right)}{b}\left[\frac{\cos \theta-\cos \phi}{\cot \theta(a) \cos \phi-\cot \phi a \cos \theta}\right]\end{aligned}$
$\begin{aligned} & \Rightarrow y=\frac{-\left(a^2+b^2\right)}{b} \\ & {\left[\frac{\cos \theta-\cos \left(\frac{\pi}{2}-\theta\right)}{\cot \theta \cos \left(\frac{\pi}{2}-\theta\right)-\cot \left(\frac{\pi}{2}-\theta\right) \cos \theta}\right]} \\ & \Rightarrow y=\frac{-a^2+b^2}{b}\left[\frac{\cos \theta-\sin \theta}{\cot \theta \sin \theta-\tan \theta \cos \theta}\right]\end{aligned}$
$\begin{aligned} & \Rightarrow \quad y=\frac{-\left(a^2+b^2\right)}{b}\left[\frac{\cos \theta-\sin \theta}{\frac{\cos \theta}{\sin \theta} \cdot(\sin \theta)-\frac{\sin \theta}{\cos \theta} \cdot \cos \theta}\right] \\ & \Rightarrow \quad y=\frac{-\left(a^2+b^2\right)}{b}\left[\frac{\cos \theta-\sin \theta}{\cos \theta-\sin \theta}\right] \\ & \Rightarrow \quad y=-\left(\frac{a^2+b^2}{b}\right)\end{aligned}$
As $(h, k)$ is the point of intersection,
So, $\quad k=-\left(\frac{a^2+b^2}{b}\right)$
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The equation of the normal at a point $P(a \sec \theta, b \tan \theta)$ is given $\frac{b y}{\tan \theta}+\frac{a x}{\sec \theta}=a^2+b^2$ $\Rightarrow \quad b \cot \theta y+a \cos \theta x-\left(a^2+b^2\right)=0$
The equation of normal at a point $\theta(a \sec \phi, b \tan \phi)$ is given
$\begin{array}{r}\frac{b y}{\tan \phi}+\frac{a x}{\sec \phi}=a^2+b^2 \\ \Rightarrow b \cot \phi y+a \cos \phi x-\left(a^2+b^2\right)=0\end{array}$
Solving Eqs. (i) and (ii), we get
$\begin{aligned} & \frac{x}{-\left(a^2+b^2\right) b \cot \theta+\left(a^2+b^2\right) b \cot \phi} \\ & =\frac{-y}{-\left(a^2+b^2\right) a \cos \phi+\left(a^2+b^2\right) a \cos \theta} \\ & =\frac{1}{b \cot \theta a \cos \phi-b \cot \phi a \cos \theta} \\ & \Rightarrow \frac{-y}{-\left(a^2+b^2\right) a \cos \phi+\left(a^2+b^2\right) a \cos \theta} \\ & =\frac{1}{b \cot \theta a \cos \phi-b \cot \phi a \cos \theta} \\ & \Rightarrow \frac{-y}{a\left(a^2+b^2\right)[\cos \theta-\cos \phi} \\ & =\frac{1}{a b[\cot \theta(a) \cos \phi-\cos \phi(a) \cos \theta} \\ & \Rightarrow y=\frac{-\left(a^2+b^2\right)}{b}\left[\frac{\cos \theta-\cos \phi}{\cot \theta(a) \cos \phi-\cot \phi a \cos \theta}\right]\end{aligned}$
$\begin{aligned} & \Rightarrow y=\frac{-\left(a^2+b^2\right)}{b} \\ & {\left[\frac{\cos \theta-\cos \left(\frac{\pi}{2}-\theta\right)}{\cot \theta \cos \left(\frac{\pi}{2}-\theta\right)-\cot \left(\frac{\pi}{2}-\theta\right) \cos \theta}\right]} \\ & \Rightarrow y=\frac{-a^2+b^2}{b}\left[\frac{\cos \theta-\sin \theta}{\cot \theta \sin \theta-\tan \theta \cos \theta}\right]\end{aligned}$
$\begin{aligned} & \Rightarrow \quad y=\frac{-\left(a^2+b^2\right)}{b}\left[\frac{\cos \theta-\sin \theta}{\frac{\cos \theta}{\sin \theta} \cdot(\sin \theta)-\frac{\sin \theta}{\cos \theta} \cdot \cos \theta}\right] \\ & \Rightarrow \quad y=\frac{-\left(a^2+b^2\right)}{b}\left[\frac{\cos \theta-\sin \theta}{\cos \theta-\sin \theta}\right] \\ & \Rightarrow \quad y=-\left(\frac{a^2+b^2}{b}\right)\end{aligned}$
As $(h, k)$ is the point of intersection,
So, $\quad k=-\left(\frac{a^2+b^2}{b}\right)$
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