QJEEM20Mac3bc: JEE Main 2020 Mathematics - Class 12 JEEM20Mac3bc Solution: If where a > b > 0, then is :

Question

Verified Solution • 2025–26 Session moderate

If

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

where a > b > 0, then

{{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)

is :

Accepted Answer

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

\Rightarrow {a^2} - \sqrt 2 ab\cos y + \sqrt 2 ab\cos x - 2{b^2}\cos x\cos y = {a^2} - {b^2}

Differentiating both sides :

0 - \sqrt 2 ab\left( { - \sin y{{dy} \over {dx}}} \right) + \sqrt 2 ab( - \sin x)

- 2{b^2}\left[ {\cos x\left( { - \sin y{{dy} \over {dx}}} \right) + \cos y( - \sin x)} \right] = 0

At

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

:

ab{{dy} \over {dx}} - ab - 2{b^2}\left( { - {1 \over 2}{{dy} \over {dx}} - {1 \over 2}} \right) = 0

\Rightarrow {{dx} \over {dy}} = {{ab + {b^2}} \over {ab - {b^2}}} = {{a + b} \over {a - b}}

; a, b > 0

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