Let P(a sec theta, b tan theta) and Q(a 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

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

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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 $ax \cos 	heta + by \cot 	heta = a^2 + b^2$.For point $P$,the normal is $ax \cos 	heta + by \cot 	heta = a^2 + b^2$.For point $Q$,the normal is $ax \cos \phi + by \cot \phi = a^2 + b^2$.Given $\phi = \frac{\pi}{2} - 	heta$,we have $\cos \phi = \sin 	heta$ and $\cot \phi = 	an 	heta$.So,the second normal is $ax \sin 	heta + by 	an 	heta = a^2 + b^2$.Subtracting the two equations: $ax(\cos 	heta - \sin 	heta) + by(\cot 	heta - 	an 	heta) = 0$.$ax(\cos 	heta - \sin 	heta) + by\left(\frac{\cos^2 	heta - \sin^2 	heta}{\sin 	heta \cos 	heta}ight) = 0$.$ax(\cos 	heta - \sin 	heta) + by\frac{(\cos 	heta - \sin 	heta)(\cos 	heta + \sin 	heta)}{\sin 	heta \cos 	heta} = 0$.Since $\cos 	heta 
eq \sin 	heta$,we divide by $(\cos 	heta - \sin 	heta)$ to get $ax + by\frac{\cos 	heta + \sin 	heta}{\sin 	heta \cos 	heta} = 0$,which implies $x = -y\frac{\cos 	heta + \sin 	heta}{a \sin 	heta \cos 	heta} \cdot \frac{b}{a}$ (simplified).Solving the system for $k$ yields $k = -\left(\frac{a^2+b^2}{b}ight)$.

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ where $\theta + \phi = \frac{\pi}{2}$ 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 drawn at $P$ and $Q$,then $k=$

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