Question

Let $P\left(\right.3sec\theta ,2tan\theta \left.\right)$ and $Q\left(\right.3sec\phi,2tan\phi\left.\right)$ where $\theta +\phi=\frac{\pi }{2},$ be two distinct points on the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1.$ Then twice the absolute value of the ordinate of the point of intersection of the normals at $P$ and $Q$ is

Answer 1

Let the coordinate at point of intersection of normals at $P$ and $Q$ be $\left(\right.h,k\left.\right)$ since, equation of normals to the hyperbola
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ At point $\left(x_{1} , y_{1}\right)$ is $\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}$
$=a^{2}+b^{2}$ therefore equation of normal to
the hyperbola $\frac{x^{2}}{3^{2}}-\frac{y^{2}}{2^{2}}=1$ at point $P\left(3 sec\theta ,2tan\theta\right)$ is
$\frac{3^{2} x}{3 sec \theta }+\frac{2^{2} y}{2 tan \theta }=3^{2}+2^{2}$
$\Rightarrow $ $3xcos\theta +2ycot\theta =3^{2}+2^{2}....\left(1\right)$
Similarly, Equation of normal to the
hyperbola $\frac{x^{2}}{3^{2}}-\frac{y^{2}}{2^{2}}$ at point $Q\left(\right.3sec\phi$ , $2tan\phi$ ) is
$\frac{3^{2} x}{3 sec \phi}+\frac{2^{2} y}{2 tan \phi}=3^{2}+2^{2}$
$\Rightarrow $ $3xcos\phi+2ycot\phi=3^{2}+2^{2}.....\left(2\right)$
Given $\theta +\phi=\frac{\pi }{2}\Rightarrow \phi=\frac{\pi }{2}-\theta $ and these
passes through $\left(\right.h,k\left.\right)$
$\therefore $ From eq. $\left(\right.2\left.\right)$
$3xcos\left(\frac{\pi }{2} - \theta \right)+2ycot\left(\frac{\pi }{2} - \theta \right)=3^{2}+2^{2}$
$\Rightarrow 3hsin\theta +2ktan\theta =3^{2}+2^{2}....\left(3\right)$
and $3hcos\theta +2kcot\theta =3^{2}+2^{2}.....\left(4\right)$
Comparing equation $\left(\right.3\left.\right)$ $\&$ $\left(\right.4\left.\right)$ , we get
$3hcos\theta +2kcot\theta =3hsin\theta +2ktan\theta $
$3hcos\theta -3hsin\theta =2ktan\theta -2kcot\theta $
$3h\left(\right.cos\theta -sin\theta \left.\right)=2k\left(\right.tan\theta -cot\theta \left.\right)$
$3h\left(\right.cos\theta -sin\theta \left.\right)$
$=2k\frac{\left(\right. sin \theta - cos \theta \left.\right) \left(\right. sin \theta + cos \theta \left.\right)}{sin \theta cos \theta }$
or, $3h=\frac{- 2 k \left(\right. sin \theta + cos \theta \left.\right)}{sin \theta cos \theta }....\left(5\right)$
Now, putting the value of equation $\left(\right.5\left.\right)$ in eq. $\left(\right.3\left.\right)$
$\frac{- 2 k \left(\right. sin \theta + cos \theta \left.\right) sin \theta }{\not{sin} cos \theta }+2ktan\theta =3^{2}+2^{2}$ $\Rightarrow 2k\not{tan \theta }$ $-2k+2k\not{tan}$ $\theta $ $=13$
$-2k=13\Rightarrow k=\frac{- 13}{2}$
Hence, ordinate of point of intersection
of normals at $P$ and $Q$ is $\frac{- 13}{2}$
So, twice the absolute value is $13$