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

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(B) 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

Vedclass · Accepted Answer

$-\frac{13}{2}$

Vedclass · Answer

(B) 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 $a^2 = 9$ and $b^2 = 4$, so $a = 3$ and $b = 2$.The normal at $P(3 \sec 	heta, 2 	an 	heta)$ is $3x \cos 	heta + 2y \cot 	heta = 9 + 4 = 13$ ... $(1)$.The normal at $Q(3 \sec \phi, 2 	an \phi)$ is $3x \cos \phi + 2y \cot \phi = 13$.Given $\phi = \frac{\pi}{2} - 	heta$, we have $\cos \phi = \sin 	heta$ and $\cot \phi = 	an 	heta$.So, the normal at $Q$ is $3x \sin 	heta + 2y 	an 	heta = 13$ ... $(2)$.From $(1)$, $3x \cos 	heta = 13 - 2y \cot 	heta \Rightarrow 3x = \frac{13 - 2y \cot 	heta}{\cos 	heta}$.From $(2)$, $3x \sin 	heta = 13 - 2y 	an 	heta \Rightarrow 3x = \frac{13 - 2y 	an 	heta}{\sin 	heta}$.Equating the two expressions for $3x$:$\frac{13 - 2y \cot 	heta}{\cos 	heta} = \frac{13 - 2y 	an 	heta}{\sin 	heta}$$(13 - 2y \cot 	heta) \sin 	heta = (13 - 2y 	an 	heta) \cos 	heta$$13 \sin 	heta - 2y \cos 	heta = 13 \cos 	heta - 2y \sin 	heta$$2y \sin 	heta - 2y \cos 	heta = 13 \cos 	heta - 13 \sin 	heta$$2y(\sin 	heta - \cos 	heta) = -13(\sin 	heta - \cos 	heta)$Since $	heta 
eq \frac{\pi}{4}$ (as $0 Thus, $2y = -13 \Rightarrow y = -\frac{13}{2}$.

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