Let P (3 sec theta, 2 tan theta) and Q (3 sec | 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

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$-\frac{13}{2}$

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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 $ax \cos 	heta + by \cot 	heta = a^2 + b^2$.For the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$,we have $a=3$ and $b=2$. Thus,the normal at $P(3 \sec 	heta, 2 	an 	heta)$ is $3x \cos 	heta + 2y \cot 	heta = 13 \dots (1)$.The normal at $Q(3 \sec \phi, 2 	an \phi)$ is $3x \cos \phi + 2y \cot \phi = 13 \dots (2)$.Given $\phi = \frac{\pi}{2} - 	heta$,we have $\cos \phi = \sin 	heta$ and $\cot \phi = 	an 	heta$.Substituting these into $(2)$,we get $3x \sin 	heta + 2y 	an 	heta = 13 \dots (3)$.Let the intersection point be $(h, k)$. From $(1)$ and $(3)$:$3h \cos 	heta + 2k \cot 	heta = 13$ and $3h \sin 	heta + 2k 	an 	heta = 13$.Equating the two expressions for $13$:$3h \cos 	heta + 2k \cot 	heta = 3h \sin 	heta + 2k 	an 	heta$$3h(\cos 	heta - \sin 	heta) = 2k(	an 	heta - \cot 	heta) = 2k \left( \frac{\sin^2 	heta - \cos^2 	heta}{\sin 	heta \cos 	heta} ight) = -2k \left( \frac{\cos 	heta - \sin 	heta}{\sin 	heta \cos 	heta} ight)(\cos 	heta + \sin 	heta)$.Assuming $\cos 	heta 
eq \sin 	heta$,we get $3h = -2k \frac{(\cos 	heta + \sin 	heta)}{\sin 	heta \cos 	heta} = -2k(\sec 	heta + \csc 	heta)$.Substituting $3h$ back into $(1)$: $(-2k(\sec 	heta + \csc 	heta)) \cos 	heta + 2k \cot 	heta = 13$ $\Rightarrow -2k(1 + \cot 	heta) + 2k \cot 	heta = 13$ $\Rightarrow -2k = 13$ $\Rightarrow k = -\frac{13}{2}$.

Let $P (3 \sec \theta, 2 \tan \theta)$ and $Q (3 \sec \phi, 2 \tan \phi)$ where $\theta + \phi = \frac{\pi}{2}$,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

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