Let P(a sec theta, b tan theta) and Q(a 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 $P(a \sec 	heta, b 	an 	heta)$ is given by $ax \si

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$-\frac{a^2 + b^2}{b}$

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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 $P(a \sec 	heta, b 	an 	heta)$ is given by $ax \sin 	heta + by = (a^2 + b^2) 	an 	heta$. Similarly,the equation of the normal at $Q(a \sec \phi, b 	an \phi)$ is $ax \sin \phi + by = (a^2 + b^2) 	an \phi$. Subtracting the two equations: $ax(\sin 	heta - \sin \phi) = (a^2 + b^2)(	an 	heta - 	an \phi)$. Using $\phi = \frac{\pi}{2} - 	heta$,we have $\sin \phi = \cos 	heta$ and $	an \phi = \cot 	heta$. Substituting these into the normal equations and solving for the intersection point $(h, k)$,we find the $y$-coordinate $k$. $by = (a^2 + b^2) 	an 	heta - ax \sin 	heta$. After simplification using the intersection condition,we obtain $k = -\frac{a^2 + b^2}{b}$.

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ be two points on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $\theta + \phi = \frac{\pi}{2}$. If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$,then $k = \dots$

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