P(a sec theta, b tan theta) and Q(a sec phi, | vedclass

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(D) The equation of the normal at point $	heta$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by $ax \cos 	heta + by \cot 	heta = a

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

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(D) The equation of the normal at point $	heta$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by $ax \cos 	heta + by \cot 	heta = a^2+b^2$.For point $P(	heta)$,the normal is $ax \cos 	heta + by \cot 	heta = a^2+b^2$ (Equation $1$).For point $Q(\phi)$,the normal is $ax \cos \phi + by \cot \phi = a^2+b^2$.Since $\phi = \frac{\pi}{2} - 	heta$,we have $\cos \phi = \sin 	heta$ and $\cot \phi = 	an 	heta$.Thus,the normal at $Q$ is $ax \sin 	heta + by 	an 	heta = a^2+b^2$ (Equation $2$).Solving for the intersection $(h, k)$ by eliminating $x$ or $y$:From Equation $1$,$ax \cos 	heta = a^2+b^2 - by \cot 	heta$.From Equation $2$,$ax \sin 	heta = a^2+b^2 - by 	an 	heta$.Using Cramer's rule or substitution,we find the $y$-coordinate $k$:$k = \frac{(a^2+b^2)a(\cos 	heta - \sin 	heta)}{ab(\sin 	heta - \cos 	heta)} = \frac{-(a^2+b^2)}{b}$.Therefore,$k = -\left(\frac{a^2+b^2}{b}ight)$.

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

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