Question

Consider the hyperbola $H:x^2–y^2=1$ and a circle $S$ with center $N(x_2,0)$. Suppose that $H$ and $S$ touch each other at a point $P(x_1,y_1)$ with $x_1>1$ and $y_1>0$. The common tangent to $H$ and $S$ at $P$ intersects the $x-$axis at point $M$. If $(l,m)$ is the centroid of the triangle $\Delta PMN$, then the correct expression(s) is(are)

• A. $\frac{dl}{d{x}_1}=1+\frac{1}{3x_1^2}\text{for}{x}_1>1$
• B. $\frac{dm}{d{x}_1}=\frac{x_1}{3\left(\sqrt{x_1^2-1}\right)}\text{for}{x}_1>1$
• C. $\frac{dl}{d{x}_1}=1-\frac{1}{3x_1^2}\text{for}{x}_1>1$
• D. $\frac{dm}{d{y}_1}=\frac{1}{3}\text{for}{y}_1>0$

Answer 1

Answer: (B), (C), (D)

$\frac{dm}{d{y}_1}=\frac{1}{3}\text{for}{y}_1>0$


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

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