Question

A curve is such that the $x$ -intercept of the tangent drawn to it at the point $P\left(x , y\right)$ is reciprocal of the abscissa of $P.$ Then, the equation of the curve is (where, $c$ is the constant of integration and $x>1$ )

• A. $y=c\left(x^{2} - 1\right)$
• B. $y=c\left(x^{2} + 1\right)$
• C. $y=c\sqrt{x^{2} - 1}$
• D. $\sqrt{y}=c\sqrt{x - 1}$

Answer 1

Answer: (C)

The general equation of the tangent is
$Y-y=\frac{d y}{d x}\left(X - x\right)$
$\therefore x$ -intercept $=x-\frac{y}{m}$
$\Rightarrow x-\frac{y}{m}=\frac{1}{x}$
$\Rightarrow x-\frac{1}{x}=\frac{y}{m}$
Or $m=\frac{y}{\left(x - \frac{1}{x}\right)}$
$\Rightarrow \frac{d y}{d x}=\frac{y}{\left(x - \frac{1}{x}\right)}$
$\Rightarrow \frac{d y}{y}=\frac{x d x}{x^{2} - 1}$
On integrating, we get
$\displaystyle \int \frac{d y}{y}=\displaystyle \int \frac{x}{x^{2} - 1}dx$
$\Rightarrow ln y=\frac{1}{2}\displaystyle \int \frac{2 x d x}{x^{2} - 1}=\frac{1}{2}ln⁡\left|x^{2} - 1\right|+ln⁡c$
$\Rightarrow ln y=ln⁡\left[\left(\sqrt{x^{2} - 1}\right) c\right]$
$\Rightarrow y=c\cdot \sqrt{x^{2} - 1}$