Hyperbola 2 Question 9
9.
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
$(1999,2 M)$
(a) $\frac{a^{2}+b^{2}}{a}$
(b) $-\frac{a^{2}+b^{2}}{a}$
(c) $\frac{a^{2}+b^{2}}{b}$
(d) $-\frac{a^{2}+b^{2}}{b}$
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Answer:
Correct Answer: 9. (d)
Solution:
- 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
$ \begin{aligned} & y-b \tan \theta=-\frac{a}{b} \sin \theta(x-a \sec \theta) \\ & \Rightarrow \quad(a \sin \theta) x+b y=\left(a^{2}+b^{2}\right) \tan \theta \\ & \Rightarrow \quad a x+b \operatorname{cosec} \theta=\left(a^{2}+b^{2}\right) \sec \theta \end{aligned} $
Similarly, equation of normal to $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at
$(a \sec \varphi, b \tan \varphi)$ is $a x+b y \operatorname{cosec} \varphi=\left(a^{2}+b^{2}\right) \sec \varphi$
On subtracting Eq. (ii) from Eq. (i), we get
$ b(\operatorname{cosec} \theta-\operatorname{cosec} \varphi) y=\left(a^{2}+b^{2}\right)(\sec \theta-\sec \varphi) $
$\Rightarrow \quad y=\frac{a^{2}+b^{2}}{b} \cdot \frac{\sec \theta-\sec \varphi}{\operatorname{cosec} \theta-\operatorname{cosec} \varphi}$
But $\frac{\sec \theta-\sec \varphi}{\operatorname{cosec} \theta-\operatorname{cosec} \varphi}=\frac{\sec \theta-\sec (\pi / 2-\theta)}{\operatorname{cosec} \theta-\operatorname{cosec}(\pi / 2-\theta)}$
$ =\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta-\sec \theta}=-1 $
$ y=-\frac{a^{2}+b^{2}}{b} \text {, i.e. } k=-\frac{a^{2}+b^{2}}{b} $
