Q. If $x^{2} + y^{2} = t + \frac{1}{t}\ and\ x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}},\ \ then\frac{dy}{dx}\$is equal to
A) $\frac{y}{x}$
B) $- \frac{y}{x}$
C) $\frac{x}{y}$
D) $- \frac{x}{y}$
Explanation
(b)
We have, $x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}\$
=$\ \left( t + \frac{1}{t} \right)^{2} - 2$
= $\left( x^{2} + y^{2} \right)^{2} - 2$
= $x^{4} + y^{4} + 2x^{2}y^{2} - 2$
$\Rightarrow \ x^{2}y^{2} = 1$
On differentiating w.r.t. $x,\$we get
$x^{2}.2y\frac{dy}{dx} + y^{2}.2x = 0$
$\Rightarrow \frac{dy}{dx} = \ - \frac{y}{x}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.