Question

Let $P(a\sec \theta ,b\tan \theta )$ and $Q(a\sec \varphi ,b\tan \varphi )$, 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

Answer: (D)

Firstly, we obtain the slope of normal to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
at $(a\sec \theta ,b\tan \theta )$.
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 $(a\sec \theta ,b\tan \theta )$ will be
$-\frac{a^{2}b\tan \theta }{b^{2}a\sec \theta }=-\frac{a}{b}\sin \theta$
$\therefore$ Equation of normal at $(a\sec \theta ,b\tan \theta )$ is
$y-b\tan \theta =-\frac{a}{b}\sin \theta (x-a\sec \theta )$
$\Rightarrow (a\sin \theta )x+by=\left(a^2+b^2\right)\tan \theta$
$\Rightarrow ax+b\mathrm{cosec}\theta =\left(a^2+b^2\right)\sec \theta ....$(i)
Similarly, equation of normal to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at
$(a\sec \varphi ,b\tan \varphi )$ is $ax+by\mathrm{cosec}\varphi =\left(a^2+b^2\right)\sec \varphi \dots$ (ii)
On subtracting Eq. (ii) from Eq. (i), we get
$b(\mathrm{cosec}\theta -\mathrm{cosec}\varphi )y=\left(a^2+b^2\right)(\sec \theta -\sec \varphi )$
$\Rightarrow y=\frac{a^2+b^2}{b}\cdot \frac{\sec \theta -\sec \varphi }{\mathrm{cosec}\theta -\mathrm{cosec}\varphi }$
But $\frac{\sec \theta -\sec \varphi }{\mathrm{cosec}\theta -\mathrm{cosec}\varphi }=\frac{\sec \theta -\sec (\pi /2-\theta )}{\mathrm{cosec}\theta -\mathrm{cosec}(\pi /2-\theta )}$
$[\because \varphi +\theta =\pi /2]$
$=\frac{\sec \theta -\mathrm{cosec}\theta }{\sec \theta -\sec \theta }=-1$
Thus, $y=-\left(\frac{a^2+b^2}{b}\right),\text{ i.e. }k=-\left(\frac{a^2+b^2}{b}\right)$