Q65 / 150 · MHT CET 2020 (14 Oct Shift 1) · Mathematics

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

  • Correct answer
    A
    $-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
  • B
    $-\sqrt{\frac{1-x^{2}}{1-y^{2}}}$
  • C
    $\sqrt{\frac{1+y^{2}}{1+x^{2}}}$
  • D
    $\sqrt{\frac{1-x^{2}}{1-y^{2}}}$

Solution

Given $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$ Put $x=\sin \alpha$ and $y=\sin \beta$ Given equation becomes $\sin \beta \cos \alpha+\sin \alpha \cos \beta=1 \Rightarrow \sin (\alpha+\beta)=1 \Rightarrow \alpha+\beta=\sin ^{-1}(1)$ $\therefore \sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}$ Differentiating w.r.t. $x$ $$ \begin{aligned} & \frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-y^{2}} \frac{d y}{d x}}=0 \\ \therefore & \frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}=-\frac{1}{\sqrt{1-x^{2}}} \Rightarrow \frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}} \end{aligned} $$

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

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