let $P(3\sec \theta ,2\tan \theta )$ and $Q(3\sec \varphi ,2\tan \varphi )$ where $\theta +\varphi =\frac{\pi }{2}$, be two distinct points on the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$.Then the ordinate of the point of intersection of the normals at $P$ and $Q$ is 

An equation of normal at $P\text{ is }3x\cos \theta +2y\cot \theta =9+4$.

$\Rightarrow 3x+2y\mathrm{cosec}\theta ==13\sec \theta$

An equation of normal at $Q$ is

$3x+2y\mathrm{cosec}\varphi =13\sec \varphi$

Subtracting (2) from (1 ), we get 

$\begin{array}{l}2y[\mathrm{cosec}\theta -\mathrm{cosec}\varphi ]=13(\sec \theta -\sec \varphi )\\ \Rightarrow 2y[\mathrm{cosec}\theta -\sec \theta ]=13(\sec \theta -\mathrm{cosec}\theta )\\ (\varphi =\pi /2-\theta )\\ \Rightarrow y=-\frac{13}{2}\end{array}$