If (a+{2} b x)(a-{2} b y)=a²-b², where a > b > 0, then d x/d y at ≤ft(π/4, π/4) is

Correct Answer is : a+ba−ba+b/a-ba−ba+b

Correct Answer is : a+ba−ba+b/a-ba−ba+b

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MHTCET 15 May 2024 Shift 1 Paper

Question : 115 of 150

Marks: +1, -0

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

a > b > 0

\frac{d x}{d y}

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

Solution:

Concept:

Implicit differentiation of a given relation in

x

and

y

to find

\frac{dx}{dy}

at a specific point.

Explanation:

We are given:

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

Differentiate both sides with respect to

y

, using the product rule.

Let

u=a+\sqrt{2}b\cos x

and

v=a-\sqrt{2}b\cos y

Then

u'=-\sqrt{2}b\sin x \cdot \frac{dx}{dy}

and

v'=\sqrt{2}b\sin y

Applying the product rule:

(a-\sqrt{2}b\cos y)\left(-\sqrt{2}b\sin x \frac{dx}{dy}\right) + (a+\sqrt{2}b\cos x)(\sqrt{2}b\sin y)=0

Solving for

\frac{dx}{dy}

\frac{dx}{dy}=\frac{\sqrt{2}b\sin y(a+\sqrt{2}b\cos x)}{\sqrt{2}b\sin x(a-\sqrt{2}b\cos y)}

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

, we have

\sin x=\sin y=\cos x=\cos y=\frac{1}{\sqrt{2}}

So the factor

\frac{\sqrt{2}b\sin y}{\sqrt{2}b\sin x}=1

, and the expression becomes:

\frac{dx}{dy}=\frac{a+\sqrt{2}b\cdot \frac{1}{\sqrt{2}}}{a-\sqrt{2}b\cdot \frac{1}{\sqrt{2}}}=\frac{a+b}{a-b}

Answer:

\frac{a+b}{a-b}

Hence, the correct option is B.

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