Question

If$(a+\sqrt{2}b\cos x)(a-\sqrt{2}b\cos y)=a^2-b^2$, where $a>b>0$, then $dy/dx$ at $(\pi /4,\pi /4)$ is

• A. $\frac{(a+b)}{(a-b)}$
• B. $\frac{(a-2b)}{(a+2b)}$
• C. $\frac{(a-b)}{(a+b)}$
• D. $\frac{(2a+b)}{(2a-b)}$

Answer 1

Answer: (A), (C)

$\frac{(a-b)}{(a+b)}$

Explanation of the correct option :

Given $(a+\sqrt{2}b\cos x)(a-\sqrt{2}b\cos y)=a^2-b^2$

Differentiating both sides with respect to $y$. we get

$-\sqrt{2}b\sin x\left(\frac{dx}{dy}\right)(a-\sqrt{2}b\cos y)+(a+\sqrt{2}b\cos x)(\sqrt{2}b\sin y)=0$

Now, $(x,y)=(\frac{\pi }{4},\frac{\pi }{4})$

$$\begin{gathered} \Rightarrow -b\left(\frac{dx}{dy}\right)(a-b)+(a+b)b=0 \\ \Rightarrow \left(\frac{dx}{dy}\right)=\frac{(a+b)}{(a-b)} \\ \Rightarrow \left(\frac{dy}{dx}\right)=\frac{(a-b)}{(a+b)} \end{gathered}$$

Hence, the correct option is C.