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
$3 h \cos \theta+2 k \cot \theta=3 h \sin \theta+2 k \tan \theta$
$3 h \cos \theta-3 h \sin \theta=2 k \tan \theta-2 k \cot \theta$
$3 h(\cos \theta-\sin \theta)=2 k(\tan \theta-\cot \theta)$
$3 h(\cos \theta-\sin \theta)$
$=2 k \frac{(\sin \theta-\cos \theta)(\sin \theta+\cos \theta)}{\sin \theta \cos \theta}$
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 $\frac{-13}{2}$