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:

  1. (1) $\frac{a-2 b}{a+2 b}$

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

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

  4. (4) $\frac{2 a+b}{2 a-b}$

Correct Option: , 3


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

Differentiating both sides,

$(-\sqrt{2} b \sin x)(a-\sqrt{2} b \cos y)+(a+\sqrt{2} b \cos x)$

$(\sqrt{2} b \sin y) \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=\frac{(\sqrt{2} b \sin x)(a-\sqrt{2} b \cos y)}{(a+\sqrt{2} b \cos x)(\sqrt{2} b \sin y)}$

$\therefore\left[\frac{d y}{d x}\right]\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$

$=\frac{a-b}{a+b} \Rightarrow \frac{d x}{d y}=\frac{a+b}{a-b}$

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