Question

If $\sqrt{(1-x^2)}+\sqrt{(1-y^2)}=a(x-y)$, then $\frac{dy}{dx}=$

• A. $\frac{\sqrt{1-x^2}}{\sqrt{1-y^2}}$
• B. $\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$
• C. $\frac{\sqrt{x^2-1}}{\sqrt{y^2-1}}$
• D. none of these

Answer 1

The correct option is B

$\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$

Explanation for the correct option:

Step 1: Separate constant terms with variables, then differentiate.

Given $\sqrt{(1-x^2)}+\sqrt{(1-y^2)}=a(x-y).........\left(i\right)$,

Let $x=\sin m$ and $y=\sin n$, then

$\begin{array}{l}m={\sin }^{-1}x\\ n={\sin }^{-1}y\end{array}$

Put $x=\sin m$ and $y=\sin n$ in the given equation, then it becomes

$\sqrt{(1-{\sin }^{2}m)}+\sqrt{(1-{\sin }^{2}n)}=a(\sin m-\sin n)..........\left(ii\right)$

Use the formula $\sqrt{1-{\sin }^{2}x}=\cos x$ and $\sin x-\sin y=2\cos \frac{x+y}{2}\cdot \sin \frac{x-y}{2}$ in the above equation, then

$\begin{array}{c}\Rightarrow \sqrt{(1-{\sin }^{2}m)}+\sqrt{(1-{\sin }^{2}n)}=a(\sin m-\sin n)\\ \Rightarrow \cos m+\cos n=a\left(2\cos \frac{m+n}{2}\cdot \sin \frac{m-n}{2}\right)\\ \Rightarrow 2\cos \left(\frac{m+n}{2}\right)\cdot \cos \left(\frac{m-n}{2}\right)=a\cdot 2\cos \left(\frac{m+n}{2}\right)\cdot \sin \left(\frac{m-n}{2}\right)\\ \Rightarrow \cos \left(\frac{m-n}{2}\right)=a\sin \left(\frac{m-n}{2}\right)\\ \Rightarrow \frac{\cos \left(\frac{m-n}{2}\right)}{\sin \left(\frac{m-n}{2}\right)}=a\\ \Rightarrow \cot \frac{m-n}{2}=a\\ \Rightarrow \left(\frac{m-n}{2}\right)={\cot }^{-1}a\\ \Rightarrow m-n=2{\cot }^{-1}a......\left(iii\right)\end{array}$

Step 2: Substitute back $\mathit{m}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{,}\mathbf{}\mathit{n}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{y}$ in the equation $\left(iii\right),$obtain

${\sin }^{-1}x-{\sin }^{-1}y=2{\cot }^{-1}a$

Then, differentiate it with respect to $x$,

$\begin{array}{c}\Rightarrow \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0\\ \Rightarrow \frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\\ \Rightarrow \frac{dy}{dx}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\end{array}$

Hence, the correct option is (B).