Q69 / 150 · MHT CET 2023 (09 May Shift 1) · Mathematics

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

  • A
    $\frac{y}{x}$
  • Correct answer
    B
    $\frac{-y}{x}$
  • C
    $\frac{x}{y}$
  • D
    $\frac{-x}{y}$

Solution

$x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}$ Squaring on both sides, we get $x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}+2$ $\begin{aligned} \therefore \quad \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 x^2 y^2=\mathrm{t}^2 & +\frac{1}{\mathrm{t}^2}+2 \\ & \ldots\left[\because x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}, \text { given }\right]\end{aligned}$ $\begin{aligned} & 2 x^2 y^2=2 \\ & x^2 y^2=1 \end{aligned}$ differentiating w.r.t. $x$, we get $\begin{aligned} & x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x y^2=0 \\ & x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 x y^2 \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-2 x y^2}{2 x^2 y}=\frac{-y}{x} \end{aligned}$

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

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