Question

If ${\sin }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a$ then $\frac{dy}{dx}$

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

Answer 1

The correct option is B $\frac{y}{x}$

$\frac{x^2-y^2}{x^2+y^2}=\sin \left(\log a\right)=\lambda ,$ $[\because \sin \left(\log a\right)$ is constant $]$

$x^2-y^2=\lambda (x^2+y^2)=\lambda x^2+\lambda y^2$

$\therefore x^2(1-\lambda )=y^2(1+\lambda )$ ...(1)

$\Rightarrow 2x(1-\lambda )=2y\frac{dy}{dx}(1+\lambda )$ ...(2)

Dividing (2) by (1), we have

$\frac{2}{x}=\frac{2}{y}\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{y}{x}$