Consider the hyperbola H   : x 2 - y 2 = 1 and a circle S with center N x 2 ,   0 . Suppose that H…

1. $\frac{dl}{d{x}_1}=1- \frac{1}{3x_1^2} for x_1>1$
2. $\frac{dm}{d{x}_1}= \frac{x_1}{3\left(\sqrt{x_1^2-1}\right)} for x_1>1$
3. $\frac{dl}{d{x}_1}=1+ \frac{1}{3x_1^2} for x_1>1$
4. $\frac{dm}{d{y}_1}=\frac{1}{3} for y_1>0$

Solution

Equation of tangent at P on hyperbola.
$x{x}_1-y{y}_1=1$
Point $M\left(\frac{1}{x_1} , 0\right)$
Equation of normal at P
$\frac{x}{x_1}+\frac{y}{y_1}=2$
Since $(x_2,0)$ satisfies its
$x_2=2x_1$
Centroid $\left(l,m\right) \equiv \left(x_1+\frac{1}{3x_1}, \frac{y_1}{3}\right)$
$\frac{dl}{d{x}_1}=1-\frac{1}{{3x}_1^2}$
$\frac{dm}{d{y}_1}=\frac{1}{3}$
$x_1^2-y_1^2=1$
$y_1=\sqrt{x_1^2-1}$
$m=\frac{\sqrt{x_1^2-1}}{3}$
$\frac{dm}{d{x}_1}=\frac{x_1}{3\sqrt{x_1^2-1}}$
Alternative Method


$l=\frac{3\sec \theta +\cos \theta }{3}$
$m=\frac{\tan \theta }{3}=\frac{y_1}{3}$
$\frac{dl}{d{x}_1}=\frac{3\sec \theta \tan \theta -\sin \theta }{3\sec \theta \tan \theta }=1-\frac{1}{3 {sec}^2 \theta }=1-\frac{1}{{3 x}_1^2}$
$\frac{dm}{d{x}_1}=\frac{\frac{{\sec }^2\theta }{3}}{\sec \theta \tan \theta }=\frac{cosec \theta }{3}=\frac{x_1}{3\sqrt{x_1^2-1}}$
$\frac{dm}{d{y}_1}=\frac{1}{3}$

Asked in: JEE Advanced 2015 (Paper 2)

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