If $(a + \sqrt{2}\, b\cos x)(a - \sqrt{2}\, b\cos y) = a^2 - b^2$, where $a > b > 0$, then $\frac{dx}{dy}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ is :

JEE Main 2020 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 4 September 2020, Shift 1, Maths Q59

Question 59 of 75 in this shift, Maths question 9 of 25.

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

, where

a > b > 0

, then

\frac{dx}{dy}

\left(\frac{\pi}{4}, \frac{\pi}{4}\right)

is :

Official answer

Option 3

NTA final key (Sep 2020).