If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=2 a^2-3 b^2$, where $ab0$, then $\frac{d x}{d y}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ is: [JEE Main 2020, 4 Sep (Shift 1)]

Question

If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=2 a^2-3 b^2$, where $a>b>0$, then $\frac{d x}{d y}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ is: [JEE Main 2020, 4 Sep (Shift 1)]

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If \((a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=2 a^2-3 b^2\), where \(a>b>0\), then \(\frac{d x}{d y}\) at \(\lef…

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If \((a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=2 a^2-3 b^2\), where \(a&gt;b&gt;0\), then \(\frac{d x}{d y}\) at \(\lef…

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If \((a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=2 a^2-3 b^2\), where \(a>b>0\), then \(\frac{d x}{d y}\) at \(\lef…