Let $P(a\sec{ heta,\ b an{ heta)}}$ and $Q(a\sec{\phi,\ b an{\phi),}}$ where $ heta + \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 normals at $P\ and\ Q$, then $k$ is equal to

Question

Let $P(a\sec{	heta,\ b	an{	heta)}}$ and $Q(a\sec{\phi,\ b	an{\phi),}}$ where $	heta + \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 normals at $P\ and\ Q$, then $k$ is equal to

Dr. Abubacker · Accepted Answer

(d)
Equation of the tangents at $P(a\sec{	heta,\ b	an{	heta)}}$ is
$\frac{x}{a}\sec{	heta - \frac{y}{b}	an{	heta = 1}}$
$	herefore\$ Equation of the normal at $P$ is
$ax + b\ cosec\ 	heta\ y = \left( a^{2} + b^{2} ight)\sec	heta$ …(i)
Similarly, the equation of normal at $Q(a\sec{\phi,\ b	an{\phi)}}$ is
$ax + b\ cosec\ \phi\ y = \left( a^{2} + b^{2} ight)\sec\phi$ …(ii)
On subtracting Eq. (ii) from Eq. (i), we get
$y = \frac{a^{2} + b^{2}}{b}.\frac{\sec{	heta - \sec\phi}}{cosec\ 	heta - cosec\ \phi}$
So that $k = y = \frac{a^{2} + b^{2}}{b}.\frac{\sec{	heta - \sec\left( \frac{\pi}{2} - 	heta ight)}}{cosec\ 	heta - cosec\ \left( \frac{\pi}{2} - 	heta ight)}$
$= \frac{a^{2} + b^{2}}{b} \bullet \frac{\sec{	heta - {cosec}	heta}}{cosec\ 	heta - sec\ 	heta}$
$= - \left\lbrack \frac{a^{2} + b^{2}}{b} ightbrack$

Q. 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 normals at $P\ and\ Q$, then $k$ is equal to

Explanation

Reviewed by Dr. Abubacker

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