Question

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. $-\frac{1}{x^{3}y}$

Answer 1

The correct option is D

$-\frac{1}{x^{3}y}$

Explanation for the correct option:

Eliminating parameter t:

$x^2+y^2=t+\frac{1}{t}$

By squaring both sides, we get

$\begin{array}{rcl}{x}^4+y^4+2x^2{y}^2& =& t^2+\frac{1}{t^2}+2\\ & \Rightarrow & t^2+\frac{1}{t^2}+2x^2{y}^2=t^2+\frac{1}{t^2}+2\left[\mathrm{Given}{x}^4+y^4=t^2+\frac{1}{t^2}\right]\\ \Rightarrow y^2& =& \frac{1}{x^2}\end{array}$

By differentiating w.r.t $x$, we get

$\begin{array}{rcl}2y\frac{dy}{dx}& =& \frac{-2}{x^3}\\ & \Rightarrow & \frac{dy}{dx}=-\frac{1}{x^{3}y}\end{array}$

Hence, option D is correct.