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
$Y-y=\frac{dy}{dx}\left(X-x\right)$
$\therefore x$ -intercept $=x-\frac{y}{m}$
$\Rightarrow x-\frac{y}{m}=\frac{1}{x}$
$\Rightarrow x-\frac{1}{x}=\frac{y}{m}$
Or $m=\frac{y}{\left(x-\frac{1}{x}\right)}$
$\Rightarrow \frac{dy}{dx}=\frac{y}{\left(x-\frac{1}{x}\right)}$
$\Rightarrow \frac{dy}{y}=\frac{xdx}{x^2-1}$
On integrating, we get
$\int \frac{dy}{y}=\int \frac{x}{x^2-1}dx$
$\Rightarrow lny=\frac{1}{2}\int \frac{2xdx}{x^2-1}=\frac{1}{2}ln\left|x^2-1\right|+lnc$
$\Rightarrow lny=ln\left[\left(\sqrt{x^2-1}\right)c\right]$
$\Rightarrow y=c\cdot \sqrt{x^2-1}$