JEE Main 2021 — Hyperbola Question with Solution

We have,

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 ...\left(i\right)$

On differentiating w.r.t., $x,$ we get

$\frac{2x}{a^2}-\frac{2y}{b^2}\times \frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{b^2}{a^2}\left(\frac{x}{y}\right)$

$\Rightarrow -\frac{dx}{dy}=-\frac{a^2}{b^2}\left(\frac{y}{x}\right)$

Slope for normal at the point $P\left(a\sec \theta ,b\tan \theta \right)$ is

${\left(-\frac{dx}{dy}\right)}_P=-\frac{a^{2}b\tan \theta }{b^{2}a\sec \theta }=-\frac{a}{b}\sin \theta$

$\therefore$ Equation of normal at $\left(a\sec \theta ,b\tan \theta \right)$ is

$y-b\tan \theta =-\frac{a}{b}\sin \theta \left(x-a\sec \theta \right)$

$\Rightarrow \left(a\sin \theta \right)x+by=\left(a^2+b^2\right)\tan \theta$

$\Rightarrow ax+bcosec\theta y=\left(a^2+b^2\right)\sec \theta ...\left(ii\right)$

Similarly, equation of normal to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at

$(a\sec \varphi ,b\tan \varphi )$ is

$ax+bycosec\varphi =\left(a^2+b^2\right)\sec \varphi ...\left(ii\right)$

On subtracting $\left(ii\right)$ from $\left(i\right)$, we get

$b(cosec\theta -cosec\varphi )y=\left(a^2+b^2\right)(\sec \theta -\sec \varphi )$

$\Rightarrow y=\frac{a^2+b^2}{b}·\left(\frac{\sec \theta -\sec \varphi }{cosec\theta -cosec\varphi }\right)$

But $\frac{\sec \theta -\sec \varphi }{cosec\theta -cosec\varphi }=\frac{\sec \theta -\sec \left({\displaystyle \frac{\pi }{2}}-\theta \right)}{cosec\theta -cosec\left({\displaystyle \frac{\pi }{2}}-\theta \right)}$

$\left[\because \varphi +\theta =\frac{\pi }{2}\right]$

$=\frac{\sec \theta -cosec\theta }{\mathrm{cosec}\theta -\sec \theta }=-1$

Thus,

$y=-\left(\frac{a^2+b^2}{b}\right),$ i.e., $k=2$.