Let P(3 θ, 2 θ) and Q(3 φ, 2 φ) where θ+φ=(π)/(2) , be two distinct...

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$$\begin{aligned}&\frac{3^2 x}{3 \sec \theta}+\frac{2^2 y}{2 \tan \theta}=3^2+2^2 \\ &\Rightarrow 3 x \cos \theta+2 y \cot \theta=3^2+2^2 \end{aligned}$$Similarly, Equation$ of normal to the hyperbola\frac{x^2}{3^2}-\frac{y^2}{2^2}$at point Q$(3 \sec \phi, 2 \tan \phi)$is$$\begin{aligned}&\frac{3^2 x}{3 \sec \phi}+\frac{2^2 y}{2 \tan \phi}=3^2+2^2 \\ &\Rightarrow 3 x \cos \phi+2 y \cot \phi=3^2+2^2 \end{aligned}$$Given$\theta+\phi=\frac{\pi}{2} \Rightarrow \phi=\frac{\pi}{2}-\theta$and these passes through$(h, k)\therefore$From eq. (2)$$\begin{aligned}&3 x \cos \left(\frac{\pi}{2}-\theta\right)+2 y \cot \left(\frac{\pi}{2}-\theta\right)=3^2+2^2 \\ &\Rightarrow 3 h \sin \theta+2 k \tan \theta=3^2+2^2 \end{aligned}$$and 3 h \cos \theta+2 k \cot \theta=3^2+2^2$ Comparing equation (3) \& (4), we get $3 h \cos \theta+2 k \cot \theta=3 h \sin \theta+2 k \tan \theta3 h \cos \theta-3 h \sin \theta=2 k \tan \theta-2 k \cot \theta3 h(\cos \theta-\sin \theta)=2 k(\tan \theta-\cot \theta)3 h(\cos \theta-\sin \theta)=2 k \frac{(\sin \theta-\cos \theta)(\sin \theta+\cos \theta)}{\sin \theta \cos \theta}$ or, $3 \mathrm{~h}=\frac{-2 k(\sin \theta+\cos \theta)}{\sin \theta \cos \theta}$ Now, putting the value of equation (5) in eq. (3) $$\begin{aligned}&\frac{-2 k(\sin \theta+\cos \theta) \sin \theta}{\sin \theta \cos \theta}+2 k \tan \theta=3^2+2^2 \\ &\Rightarrow 2 k \tan \theta-2 k+2 k \tan \theta=13 \\ &-2 k=13 \Rightarrow k=\frac{-13}{2} \end{aligned}$$Hence, ordinate of point of intersection of normals at $P and Q$ is $\frac{-13}{2}$