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: , 2
$(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}$
$\Rightarrow a^{2}-\sqrt{2} a b \cos y+\sqrt{2} a b \cos x$
$-2 b^{2} \cos x \cos y=a^{2}-b^{2}$
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 $\left(\frac{\pi}{4}, \frac{\pi}{4}\right):$
$\mathrm{ab} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{ab}-2 \mathrm{~b}^{2}\left(-\frac{1}{2} \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{2}\right)=0$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{ab}+\mathrm{b}^{2}}{\mathrm{ab}-\mathrm{b}^{2}}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}-\mathrm{b}}$ $a, b>0$
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