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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 :

1. $\frac{a-b}{a+b}$

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

3. $\frac{2 a+b}{2 a-b}$

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

Correct Option: , 2

Solution:

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

$\Rightarrow a^{2}-\sqrt{2} a b \cos y+\sqrt{2} a b \cos x$

$-2 b^{2} \cos x \cos y=a^{2}-b^{2}$

Differentiating both sides :

$0-\sqrt{2} \mathrm{ab}\left(-\sin \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}\right)+\sqrt{2} \mathrm{ab}(-\sin \mathrm{x})$

$-2 b^{2}\left[\cos x\left(-\sin y \frac{d y}{d x}\right)+\cos y(-\sin x)\right]=0$

At $\left(\frac{\pi}{4}, \frac{\pi}{4}\right):$

$\mathrm{ab} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{ab}-2 \mathrm{~b}^{2}\left(-\frac{1}{2} \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{2}\right)=0$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{ab}+\mathrm{b}^{2}}{\mathrm{ab}-\mathrm{b}^{2}}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}-\mathrm{b}}$ $a, b>0$