Q71 / 150 · MHT CET 2024 (11 May Shift 1) · Mathematics

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

  • A
    $\frac{1}{x^3 y}$
  • B
    $\frac{1}{x y^3}$
  • C
    $-\frac{1}{x y^3}$
  • Correct answer
    D
    $-\frac{1}{x^3 y}$

Solution

$\begin{array}{ll} & x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}} \text { and } x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2} \\ \therefore \quad & \left(x^2+y^2\right)^2=\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^2 \\ \therefore \quad & x^4+y^4+2 x^2 y^2=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 \\ \therefore \quad & \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 x^2 y^2=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 \\ \therefore \quad & x^2 \dot{y}^2=1 \\ \therefore \quad & y^2=\frac{1}{x^2}\end{array}$ Differentiating w.r.t. $x$, we get $\begin{aligned} & 2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-2}{x^3} \\ \therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-1}{x^3 y} \end{aligned}$

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

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