Let $\mathrm{A}(2 \sec \mathrm{\theta }, 3 \tan \mathrm{\theta })$ and $\mathrm{B}(2 \sec \mathrm{\psi }, 3 \tan \mathrm{\psi })$ where $\mathrm{\theta }+\mathrm{\psi }=\frac{\mathrm{\pi }}{2},$ be two points on the hyperbola $\frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{9}=1.$ If $(\mathrm{\alpha }, \mathrm{\beta })$ is point of intersection of normals to the hyperbola at A and B then $\mathrm{\beta }=$

$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{9}=1 \\ where {\mathrm{a}}^2=4, {\mathrm{b}}^2=9 \\ \mathrm{Given} \mathrm{A}=(2\mathrm{sec\theta }, 3\mathrm{tan\theta }), \mathrm{B}=(2\mathrm{se}c\mathrm{\psi }, 3\mathrm{tan\psi }), \mathrm{\theta }+\mathrm{\psi }=\frac{\mathrm{\pi }}{2} \\ \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{at} \mathrm{A} \mathrm{is} \\ \frac{2\mathrm{x}}{\mathrm{sec\theta }}+\frac{3\mathrm{y}}{\mathrm{tan\theta }}=13 \\ \Rightarrow 2\mathrm{xcos\theta }+3\mathrm{ycot\theta }=13 — ① \\ \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{at} \mathrm{B} \mathrm{is} \\ \frac{2\mathrm{x}}{\mathrm{sec\psi }}+\frac{3\mathrm{y}}{\mathrm{tan\psi }}=13 \\ \Rightarrow 2\mathrm{x} \mathrm{cos\psi }+3\mathrm{y} \mathrm{cot\psi }=13 \\ \Rightarrow 2\mathrm{x} \mathrm{sin\theta }+3\mathrm{y} \mathrm{tan\theta }=13 — ② \left(\therefore \mathrm{\theta }+\mathrm{\psi }=\frac{\mathrm{\pi }}{2}\right) \\ Now \left(1\right)\sin \theta -\left(2\right)\cos \theta \Rightarrow 3y\cos \theta -3y\sin \theta =13\sin \theta -13\cos \theta \\ \Rightarrow 3y(\cos \theta -\sin \theta )=-13(\cos \theta -\sin \theta ) \\ \Rightarrow y=\frac{-13}{3} \\ Given point of intersection of ① \& ② is (\alpha ,\beta )\Rightarrow \beta =\frac{-13}{3} \end{gathered}$$