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 Main · 2020 · 04 Sep 2020 · Shift 1 · Derivatives XI

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

, where

a > b > 0

, then

\frac{d x}{d y}

(\frac{π}{4}, \frac{π}{4})

is:

Answer: (C)

Differentiate

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

with respect to

y

(a - 2 b cos y) (- 2 b sin x \frac{d x}{d y}) + (a + 2 b cos x) (2 b sin y) = 0

. At

(\frac{π}{4}, \frac{π}{4})

, where

sin = cos = \frac{1}{2}

, the factor

a - 2 b cos y = a - b

and

a + 2 b cos x = a + b

, so

- (a - b) b \frac{d x}{d y} + (a + b) b = 0

, giving

\frac{d x}{d y} = \frac{a + b}{a - b}

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option