Question

Let $P(3sec\theta ,2tan\theta )$ and $Q(3sec\varphi ,2tan\varphi )$ where $\theta +\varphi =\frac{\pi }{2},$ 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

Answer: 13

Let the coordinate at point of intersection of normals at $P$ and $Q$ be $(h,k)$ 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 $P\left(3sec\theta ,2tan\theta \right)$ is
$\frac{3^{2}x}{3sec\theta }+\frac{2^{2}y}{2tan\theta }=3^2+2^2$
$\Rightarrow$ $3xcos\theta +2ycot\theta =3^2+2^2....\left(1\right)$
Similarly, Equation of normal to the
hyperbola $\frac{x^2}{3^2}-\frac{y^2}{2^2}$ at point $Q(3sec\varphi$ , $2tan\varphi$ ) is
$\frac{3^{2}x}{3sec\varphi }+\frac{2^{2}y}{2tan\varphi }=3^2+2^2$
$\Rightarrow$ $3xcos\varphi +2ycot\varphi =3^2+2^2.....\left(2\right)$
Given $\theta +\varphi =\frac{\pi }{2}\Rightarrow \varphi =\frac{\pi }{2}-\theta$ and these
passes through $(h,k)$
$\therefore$ From eq. $(2)$
$3xcos\left(\frac{\pi }{2}-\theta \right)+2ycot\left(\frac{\pi }{2}-\theta \right)=3^2+2^2$
$\Rightarrow 3hsin\theta +2ktan\theta =3^2+2^2....\left(3\right)$
and $3hcos\theta +2kcot\theta =3^2+2^2.....\left(4\right)$
Comparing equation $(3)$ $\&$ $(4)$ , we get
$3hcos\theta +2kcot\theta =3hsin\theta +2ktan\theta$
$3hcos\theta -3hsin\theta =2ktan\theta -2kcot\theta$
$3h(cos\theta -sin\theta )=2k(tan\theta -cot\theta )$
$3h(cos\theta -sin\theta )$
$=2k\frac{(sin\theta -cos\theta )(sin\theta +cos\theta )}{sin\theta cos\theta }$
or, $3h=\frac{-2k(sin\theta +cos\theta )}{sin\theta cos\theta }....\left(5\right)$
Now, putting the value of equation $(5)$ in eq. $(3)$
$\frac{-2k(sin\theta +cos\theta )sin\theta }{/sincos\theta }+2ktan\theta =3^2+2^2$ $\Rightarrow 2k/tan\theta$ $-2k+2k/tan$ $\theta$ $=13$
$-2k=13\Rightarrow k=\frac{-13}{2}$
Hence, ordinate of point of intersection
of normals at $P$ and $Q$ is $\frac{-13}{2}$
So, twice the absolute value is $13$