Let A(2 sec theta, 3 tan theta) and B(2 sec p | vedclass

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(D) The equation of the normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at point $(a \sec 	heta, b 	an 	heta)$ is given by $ax \cos 	h

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$-13/3$

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(D) The equation of the normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at point $(a \sec 	heta, b 	an 	heta)$ is given by $ax \cos 	heta + by \cot 	heta = a^2+b^2$. Here $a=2$ and $b=3$, so $a^2+b^2 = 4+9 = 13$. Normal at $A(2 \sec 	heta, 3 	an 	heta)$ is: $2x \cos 	heta + 3y \cot 	heta = 13$  --- $(1)$ Since $\phi = \frac{\pi}{2} - 	heta$, we have $\sec \phi = \csc 	heta$ and $	an \phi = \cot 	heta$. Normal at $B(2 \csc 	heta, 3 \cot 	heta)$ is: $2x \sin 	heta + 3y 	an 	heta = 13$  --- $(2)$ To find the intersection $(\alpha, \beta)$, we eliminate $x$ by multiplying $(1)$ by $\sin 	heta$ and $(2)$ by $\cos 	heta$: $2x \sin 	heta \cos 	heta + 3y \cos 	heta = 13 \sin 	heta$ $2x \sin 	heta \cos 	heta + 3y \sin 	heta = 13 \cos 	heta$ Subtracting the two equations: $3y(\cos 	heta - \sin 	heta) = 13(\sin 	heta - \cos 	heta)$. Since $	heta 
eq \frac{\pi}{4}$, we can divide by $(\cos 	heta - \sin 	heta)$: $3y = -13 \implies y = -\frac{13}{3}$. Thus, $\beta = -\frac{13}{3}$.

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

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