Q. The equation of curve passing through the point $\left( 1,\frac{\pi}{4} \right)$ and having slope of tangent at any point $(x,y)$ as $\frac{y}{x} - \cos^{2}\left( \frac{y}{x} \right)$, is
A) $x = e^{1 + \tan\left( \frac{y}{x} \right)}$
B) $x = e^{1 - \tan\left( \frac{y}{x} \right)}$
C) $x = e^{1 + \tan\left( \frac{x}{y} \right)}$
D) $x = e^{1 - \tan\left( \frac{x}{y} \right)}$
Explanation
(b)
Given, $\frac{x\ dy - y\ dx}{x} = - \left( \cos^{2}\frac{y}{x} \right)dx$
$\Rightarrow$ $\sec^{2}{\left( \frac{y}{x} \right)\left( \frac{x\ dy - y\ dx}{x^{2}} \right) = - \frac{dx}{x}}$
$\Rightarrow$ $\sec^{2}\left( \frac{y}{x} \right)d\left( \frac{y}{x} \right) = - \frac{dx}{x}$
$\Rightarrow$ $\tan\frac{y}{x} = - \log x + c$ [integrating]
When $x = 1,\ \ y = \frac{\pi}{4}\ \ \ \Rightarrow c = 1$
$\therefore$ $\tan\left( \frac{y}{x} \right) = 1 - \log x\ \ \ \ \Rightarrow x = e^{1 - \tan\left( \frac{y}{x} \right)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.