If $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$, then $\frac{d y}{d x}$ is equal to

• A

  $-\frac{x}{y}$
• Correct answer
  B

  $-\frac{y}{x}$
• C

  $-\frac{x^2}{y^2}$
• D

  $-\frac{y^2}{x^2}$

Solution

We have,
From Eq. (i), squaring both the sides $\begin{aligned} & \left(x^2+y^2\right)^2=\left(t+\frac{1}{t}\right)^2 \\ & x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}+2 \\ & \left.x^4+y^4+2 x^2 y^2=x^4+y^4+2 \quad \text { [using Eq. (ii) }\right] \\ & 2 x^2 y^2=2 \end{aligned}$ Differentiating w.r.t. $x$, we get $\begin{aligned} & 2 x y^2+2 x^2 y \frac{d y}{d x}=0 \\ & \therefore \quad \frac{d y}{d x}=\frac{-2 x y^2}{2 x^2 y}=-\frac{y}{x} \end{aligned}$

Hence, the correct answer is option: 'B'.