Q109 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

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

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

Solution

Given $\mathrm{x}^2+\mathrm{y}^2=\mathrm{t}-\frac{1}{\mathrm{t}}$ $\begin{aligned} & \text { and } x^4+y^4=t^2+\frac{1}{t^2} \\ & \Rightarrow x^4+y^4=x^4+y^4+2 x^2 y^2+2 \\ & \Rightarrow x^2 y^2+1=0 \end{aligned}$ Now $\frac{d y}{d x} \Rightarrow 2 x y^2+x^2 2 y y l=0$ $\begin{aligned} & \Rightarrow y^1=\frac{-2 x y^2}{2 x^2 y} \\ & =\frac{-y}{x} \end{aligned}$

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

Relevant Questions