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Q.
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 :
JEE MainJEE Main 2020Continuity and Differentiability
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 :
$0-\sqrt{2} a b\left(-\sin y \frac{d y}{d x}\right)+\sqrt{2} a b(-\sin x)$
$-2 b^{2}\left[\cos x\left(-\sin y \frac{d y}{d x}\right)+\cos y(-\sin x)\right]=0$
At $\left(\frac{\pi}{4}, \frac{\pi}{4}\right):$
$a b \frac{d y}{d x}-a b-2 b^{2}\left(-\frac{1}{2} \frac{d y}{d x}-\frac{1}{2}\right)=0$
$\Rightarrow \frac{d x}{d y}=\frac{a b+b^{2}}{a b-b^{2}}=\frac{a+b}{a-b} ; a, b>0$