Search any question & find its solution
Question:
Answered & Verified by Expert
If $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1, \quad$ then $\frac{d y}{d x}=$
Options:
Solution:
1309 Upvotes
Verified Answer
The correct answer is:
$-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
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. $x$
$$
\begin{aligned}
& \frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-y^{2}} \frac{d y}{d x}}=0 \\
\therefore & \frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}=-\frac{1}{\sqrt{1-x^{2}}} \Rightarrow \frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}
\end{aligned}
$$
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. $x$
$$
\begin{aligned}
& \frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-y^{2}} \frac{d y}{d x}}=0 \\
\therefore & \frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}=-\frac{1}{\sqrt{1-x^{2}}} \Rightarrow \frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.