JEE Main 2014 — Hyperbola Question with Solution

Let the coordinate at point of intersection of normals at $\mathrm{P}$ and $\mathrm{Q}$ be $(h,k)$ Since, equation of normals to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ At point $\left(x_1,y_1\right)$ is $\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$ therefore equation of normal to the hyperbola $\frac{x^2}{3_1^2}-\frac{y^2}{2^2}=1$ at point $P(3\sec \theta ,2\tan \theta )$ is $$\begin{array}{rl}& \frac{3^{2}x}{3\sec \theta }+\frac{2^{2}y}{2\tan \theta }=3^2+2^2\\ & \Rightarrow 3x\cos \theta +2y\cot \theta =3^2+2^2\end{array}$$ Similarly, Equation of normal to the hyperbola $\frac{x^2}{3^2}-\frac{y^2}{2^2}$ at point Q $(3\sec \varphi ,2\tan \varphi )$ is $$\begin{array}{rl}& \frac{3^{2}x}{3\sec \varphi }+\frac{2^{2}y}{2\tan \varphi }=3^2+2^2\\ & \Rightarrow 3x\cos \varphi +2y\cot \varphi =3^2+2^2\end{array}$$ Given $\theta +\varphi =\frac{\pi }{2}\Rightarrow \varphi =\frac{\pi }{2}-\theta$ and these passes through $(h,k)$ $\therefore$ From eq. (2) $$\begin{array}{rl}& 3x\cos \left(\frac{\pi }{2}-\theta \right)+2y\cot \left(\frac{\pi }{2}-\theta \right)=3^2+2^2\\ & \Rightarrow 3h\sin \theta +2k\tan \theta =3^2+2^2\end{array}$$ and $3h\cos \theta +2k\cot \theta =3^2+2^2$ Comparing equation (3) \& (4), we get $3h\cos \theta +2k\cot \theta =3h\sin \theta +2k\tan \theta$ $3h\cos \theta -3h\sin \theta =2k\tan \theta -2k\cot \theta$ $3h(\cos \theta -\sin \theta )=2k(\tan \theta -\cot \theta )$ $3h(\cos \theta -\sin \theta )$ $=2k\frac{(\sin \theta -\cos \theta )(\sin \theta +\cos \theta )}{\sin \theta \cos \theta }$ or, $3\mathrm{h}=\frac{-2k(\sin \theta +\cos \theta )}{\sin \theta \cos \theta }$ Now, putting the value of equation (5) in eq. (3) $$\begin{array}{rl}& \frac{-2k(\sin \theta +\cos \theta )\sin \theta }{\sin \theta \cos \theta }+2k\tan \theta =3^2+2^2\\ & \Rightarrow 2k\tan \theta -2k+2k\tan \theta =13\\ & -2k=13\Rightarrow k=\frac{-13}{2}\end{array}$$ Hence, ordinate of point of intersection of normals at $P$ and $Q$ is $\frac{-13}{2}$