Question

Let $P(asec\theta ,btan\theta )$ and $Q(asec\varphi ,btan\varphi )$, where $\theta +\varphi =\frac{\pi }{2}$, be two points on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
If (h, k) is the point of the intersection of the normals at P and Q, then k is equal to

• A. $\frac{a^2+b^2}{a}$
• B. $-\left(\frac{a^2+b^2}{a}\right)$
• C. $\frac{a^2+b^2}{b}$
• D. $-\left(\frac{a^2+b^2}{b}\right)$

Answer 1

The correct option is D

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

Firstly, we obtain the slope of normal to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at $(asec\theta ,btan\theta )$
On differentiating w.r.t.x, we get
$\frac{2x}{a^2}-\frac{2y}{b^2}\times \frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{b^2}{a^2}\frac{x}{y}$
Slope for normal at the point $(asec\theta ,btan\theta )$ will be $-\frac{a^{2}btan\theta }{b^{2}asec\theta }=-\frac{a}{b}sin\theta$
$\therefore$ Equation of normal at $(asec\theta ,btan\theta )$ is
$y-btan\theta =-\frac{a}{b}sin\theta (x-asec\theta )$
$\Rightarrow \left(asin\theta \right)x+by=(a^2+b^2)tan\theta$
$\Rightarrow ax+bycosec\theta =(a^2+b^2)sec\theta$ ... (i)
Similarly, equation of normal to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at $(asec\varphi ,btan\varphi )$ is $ax+bycosec\varphi =(a^2+b^2)sec\varphi$ ... (ii)
On subtracting Eq. (ii) from Eq. (i), we get
$b(cosec\theta -cosec\varphi )y=(a^2+b^2)(sec\theta -sec\varphi )$
$\Rightarrow y=\frac{a^2+b^2}{b}.\frac{sec\theta -sec\varphi }{cosec\theta -cosec\varphi }$
But $\frac{sec\theta -sec\varphi }{cosec\theta -cosec\varphi }=\frac{sec\theta -sec(\frac{\pi }{2}-\theta )}{cosec\theta -cosec(\frac{\pi }{2}-\theta )}$
$[\because \varphi +\theta =\frac{\pi }{2}]$
$=\frac{sec\theta -cosec\theta }{cosec\theta -sec\theta }=-1$
Thus, $y=-\left(\frac{a^2+b^2}{b}\right),i.e.k=-\left(\frac{a^2+b^2}{b}\right)$