Q106 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

If \(\sqrt{1-x^6}+\sqrt{1-y^6}=a\left(x^3-y^3\right)\), then \(y^2 \frac{d y}{d x}=\)

  • A
    \(\sqrt{\frac{1-y^6}{1-x^6}}\)
  • B
    \(x \sqrt{\frac{1-y^6}{1-x^6}}\)
  • Correct answer
    C
    \(x^2 \sqrt{\frac{1-y^6}{1-x^6}}\)
  • D
    \(\frac{1}{x^2} \sqrt{\frac{1-y^6}{1-x^6}}\)

Solution

Given equation is \(\begin{aligned} & \sqrt{1-x^6}+\sqrt{1-y^6}=a\left(x^3-y^3\right) \\ \Rightarrow \quad & \frac{\sqrt{1-x^6}+\sqrt{1-y^6}}{x^3-y^3}=a \end{aligned}\) On differentiating both sides w.r.t., \(x\), we get \(\frac{\left[\left(x^3-y^3\right)\left(\frac{-6 x^5}{2 \sqrt{1-x^6}}-\frac{6 y^5}{2 \sqrt{1-y^6}} \frac{d y}{d x}\right)-\left(\sqrt{1-x^6}+\sqrt{1-y^6}\right)\left(3 x^2-3 y^2 \frac{d y}{d x}\right)\right]}{\left(x^3-y^3\right)^2}=0\) \(\begin{aligned} & \Rightarrow\left(y^2\left(\sqrt{1-x^6}+\sqrt{1-y^6}\right)-\frac{y^5\left(x^3-y^3\right)}{\sqrt{1-y^6}}\right) \frac{d y}{d x} \\ & =x^2\left(\sqrt{1-x^6}+\sqrt{1-y^6}\right)+\frac{x^5}{\sqrt{1-x^6}}\left(x^3-y^3\right) \\ & \Rightarrow y^2 \frac{d y}{d x}\left[\frac{\sqrt{1-x^6}+\sqrt{1-y^6}+\left(1-y^6\right)-y^3 x^3+y^6}{\sqrt{1-y^6}}\right] \\ & =x^2\left[\frac{\left(1-x^6\right)+\sqrt{1-y^6} \sqrt{1-x^6}+x^6-x^3 y^3}{\sqrt{1-x^6}}\right] \\ & \Rightarrow y^2 \frac{d y}{d x}\left[\frac{\sqrt{1-x^6} \sqrt{1-y^6}+1-x^3 y^3}{\sqrt{1-y^6}}\right] \\ & =x^2\left[\frac{1+\sqrt{1-y^6} \sqrt{1-x^6}-x^3 y^3}{\sqrt{1-x^6}}\right] \\ & \left.\Rightarrow y^2 \frac{d y}{d x}=x^2 \sqrt{\frac{1-y^6}{1-x^6}}\right. \end{aligned}\) Hence, option (3) is correct.

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

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