Question:
If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}$, where $a>b>0$, then $\frac{d x}{d y}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ is:
Correct Option: , 3
Solution:
$(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}$
Differentiating both sides,
$(-\sqrt{2} b \sin x)(a-\sqrt{2} b \cos y)+(a+\sqrt{2} b \cos x)$
$(\sqrt{2} b \sin y) \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=\frac{(\sqrt{2} b \sin x)(a-\sqrt{2} b \cos y)}{(a+\sqrt{2} b \cos x)(\sqrt{2} b \sin y)}$
$\therefore\left[\frac{d y}{d x}\right]\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$
$=\frac{a-b}{a+b} \Rightarrow \frac{d x}{d y}=\frac{a+b}{a-b}$