If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$, where $\mathrm{a}\gt\mathrm{b}\gt0$, then…

If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$, where $\mathrm{a}\gt\mathrm{b}\gt0$, then $\frac{\mathrm{d} x}{\mathrm{~d} y}$ 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. $\frac{a-2 b}{a+2 b}$

Solution

$(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$
Differentiating both sides w.r.t. $y$, we get $\begin{aligned} & \begin{aligned} &(\mathrm{a}+\sqrt{2} \mathrm{~b} \cos x)(\sqrt{2} \mathrm{~b} \sin y) \\ & \quad+(\mathrm{a}-\sqrt{2} \mathrm{~b} \cos y)\left(-\sqrt{2} \mathrm{~b} \sin x \frac{\mathrm{~d} x}{\mathrm{~d} y}\right)=0 \\ & \Rightarrow \frac{\mathrm{~d} x}{\mathrm{~d} y}=\frac{\sqrt{2} \mathrm{~b} \sin y(\mathrm{a}+\sqrt{2} \mathrm{~b} \cos x)}{\sqrt{2} \mathrm{~b} \sin x(\mathrm{a}-\sqrt{2} \mathrm{~b} \cos y)} \\ & \Rightarrow\left(\frac{\mathrm{d} x}{\mathrm{~d} y}\right)_{\left(\frac{\pi}{4} \frac{\pi}{4}\right)}=\frac{\mathrm{b}(\mathrm{a}+\mathrm{b})}{\mathrm{b}(\mathrm{a}-\mathrm{b})}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}-\mathrm{b}} \end{aligned} \end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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