Let P(asec theta, btan theta) and Q(asec varp | vedclass

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

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$-\left(\frac{a^2 + b^2}{b}\right)$

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(D) The equation of the normal at $P(a\sec 	heta, b	an 	heta)$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $ax\sin 	heta + by = (a^2 + b^2)	an 	heta$. Similarly,the equation of the normal at $Q(a\sec \varphi, b	an \varphi)$ is $ax\sin \varphi + by = (a^2 + b^2)	an \varphi$. Let $(h, k)$ be the intersection point. Then $k$ satisfies: $by = (a^2 + b^2)	an 	heta - ax\sin 	heta$ and $by = (a^2 + b^2)	an \varphi - ax\sin \varphi$. Subtracting the two equations: $b(k - k) = 0$ is not the path; we solve for $y$ by eliminating $x$. From the normal equations: $ax\sin 	heta + by = (a^2 + b^2)	an 	heta$ and $ax\sin \varphi + by = (a^2 + b^2)	an \varphi$. Multiplying the first by $\sin \varphi$ and the second by $\sin 	heta$ and subtracting: $by(\sin \varphi - \sin 	heta) = (a^2 + b^2)(	an 	heta \sin \varphi - 	an \varphi \sin 	heta)$. Given $\varphi = \frac{\pi}{2} - 	heta$,so $\sin \varphi = \cos 	heta$ and $	an \varphi = \cot 	heta$. $by(\cos 	heta - \sin 	heta) = (a^2 + b^2)(	an 	heta \cos 	heta - \cot 	heta \sin 	heta)$. $by(\cos 	heta - \sin 	heta) = (a^2 + b^2)(\sin 	heta - \cos 	heta)$. $by = -(a^2 + b^2)$. Thus,$k = -\frac{a^2 + b^2}{b}$.

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

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