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. $\frac{1}{x^2{y}^2}$
• B. $\frac{1}{x^2{y}^3}$
• C. $\frac{1}{x^{3}y}$
• D. $-\frac{1}{x^{3}y}$

Answer 1

The correct option is C

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

Explanation for the correct option:

Eleminating parameter $t$:

It is given that $x^2+y^2=t-\frac{1}{t}$. By squaring both sides we get,

$x^4+y^4+2x^2{y}^2=t^2+\frac{1}{t^2}-2......\left(1\right)$

It is also given that $x^4+y^4=t^2+\frac{1}{t^2}$. By substituting the value of $x^4+y^4$ in $\left(1\right)$, we get

$\begin{array}{rcl}{t}^2+\frac{1}{t^2}+2x^2{y}^2& =& t^2+\frac{1}{t^2}-2\\ \Rightarrow 2x^2{y}^2& =& -2\\ \Rightarrow x^2{y}^2& =& -1\end{array}$

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

$x^2\left(2y\frac{d}{dx}\right)+y^2\left(2x\right)=0$

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

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

Hence, option C is correct.