Question

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$, where $\theta+\phi=\frac{\pi}{2}$, 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. $-\Bigg(\frac{a^2+b^2}{a}\Bigg)$
• C. $\frac{a^2+b^2}{b}$
• D. $-\Bigg(\frac{a^2+b^2}{b}\Bigg)$

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{2 x}{a^{2}}-\frac{2 y}{b^{2}} \times \frac{d y}{d x}=0 $
$\Rightarrow \frac{d y}{d x}=\frac{b^{2}}{a^{2}} \frac{x}{y}$
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) $
$\Rightarrow (a \sin \theta) x+b y =\left(a^{2}+b^{2}\right) \tan \theta$
$\Rightarrow a x+b \operatorname{cosec} \theta =\left(a^{2}+b^{2}\right) \sec \theta ....$(i)
Similarly, equation of normal to $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at
$(a \sec \phi, b \tan \phi)$ is $a x+b y \operatorname{cosec} \phi=\left(a^{2}+b^{2}\right) \sec \phi \ldots$ (ii)
On subtracting Eq. (ii) from Eq. (i), we get
$b(\operatorname{cosec} \theta-\operatorname{cosec} \phi) y =\left(a^{2}+b^{2}\right)(\sec \theta-\sec \phi) $
$\Rightarrow y =\frac{a^{2}+b^{2}}{b} \cdot \frac{\sec \theta-\sec \phi}{\operatorname{cosec} \theta-\operatorname{cosec} \phi}$
But $\frac{\sec \theta-\sec \phi}{\operatorname{cosec} \theta-\operatorname{cosec} \phi}=\frac{\sec \theta-\sec (\pi / 2-\theta)}{\operatorname{cosec} \theta-\operatorname{cosec}(\pi / 2-\theta)}$
$[\because \phi+\theta=\pi / 2]$
$=\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta-\sec \theta}=-1$
Thus, $y=-\left(\frac{a^{2}+b^{2}}{b}\right), \text { i.e. } k=-\left(\frac{a^{2}+b^{2}}{b}\right)$