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Question: If $(a + \sqrt{2b} \cos x)(a - \sqrt{2b} \cos y) = a^2 - b^2$, where a > b > 0, then $\frac{dx}{dy}$...

If (a+2bcos⁡x)(a−2bcos⁡y)=a2−b2(a + \sqrt{2b} \cos x)(a - \sqrt{2b} \cos y) = a^2 - b^2, where a > b > 0, then dxdy\frac{dx}{dy} at (π4,π4)(\frac{\pi}{4}, \frac{\pi}{4}) is

A

a−ba+b\frac{a-b}{a+b}

B

a+ba−b\frac{a+b}{a-b}

C

2a+b2a−b\frac{2a+b}{2a-b}

D

a−2ba+2b\frac{a-2b}{a+2b}

Answer

a+ba−b\frac{a+b}{a-b}

Explanation

Solution

We are given

(a+2bcos⁡x)(a−2bcos⁡y)=a2−b2,a>b>0.\Bigl(a+\sqrt{2b}\cos x\Bigr)\Bigl(a-\sqrt{2b}\cos y\Bigr)=a^2-b^2,\quad a>b>0.

We wish to find dxdy\frac{dx}{dy} at x=y=π4x=y=\frac{\pi}{4}.

Step 1. Differentiate implicitly with respect to yy:

Let

P=a+2bcos⁡x,Q=a−2bcos⁡y.P = a+\sqrt{2b}\cos x,\quad Q = a-\sqrt{2b}\cos y.

Differentiating PQ=a2−b2PQ=a^2-b^2 (a constant) gives:

ddy(PQ)=P′⋅Q+P⋅Q′=0.\frac{d}{dy}(PQ)=P'\cdot Q+P\cdot Q'=0.

Since x=x(y)x=x(y), we have:

P′=ddy(2bcos⁡x)=−2bsin⁡x dxdy,P'=\frac{d}{dy}\bigl(\sqrt{2b}\cos x\bigr)=-\sqrt{2b}\sin x\,\frac{dx}{dy}, Q′=ddy(−2bcos⁡y)=2bsin⁡y.Q'=\frac{d}{dy}\bigl(-\sqrt{2b}\cos y\bigr)=\sqrt{2b}\sin y.

Thus,

−2bsin⁡x (a−2bcos⁡y)dxdy+2bsin⁡y (a+2bcos⁡x)=0.-\sqrt{2b}\sin x\,(a-\sqrt{2b}\cos y)\frac{dx}{dy}+\sqrt{2b}\sin y\,(a+\sqrt{2b}\cos x)=0.

Step 2. Solve for dxdy\frac{dx}{dy}:

Divide the entire equation by 2b\sqrt{2b}:

−sin⁡x (a−2bcos⁡y)dxdy+sin⁡y (a+2bcos⁡x)=0.-\sin x\,(a-\sqrt{2b}\cos y)\frac{dx}{dy}+\sin y\,(a+\sqrt{2b}\cos x)=0.

So,

dxdy=sin⁡y (a+2bcos⁡x)sin⁡x (a−2bcos⁡y).\frac{dx}{dy}=\frac{\sin y\,(a+\sqrt{2b}\cos x)}{\sin x\,(a-\sqrt{2b}\cos y)}.

Step 3. Substitute x=y=π4x=y=\frac{\pi}{4}:

At x=y=π4x=y=\frac{\pi}{4}, we have:

sin⁡π4=cos⁡π4=12.\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}.

Thus,

dxdy=12(a+2b⋅12)12(a−2b⋅12)=a+ba−b.\frac{dx}{dy}=\frac{\frac{1}{\sqrt{2}}\Bigl(a+\sqrt{2b}\cdot\frac{1}{\sqrt{2}}\Bigr)} {\frac{1}{\sqrt{2}}\Bigl(a-\sqrt{2b}\cdot\frac{1}{\sqrt{2}}\Bigr)} =\frac{a+\sqrt{b}}{a-\sqrt{b}}.

Step 4. Matching with the given options:

The answer provided in the options has the form a±ba∓b\dfrac{a \pm b}{a \mp b}. Note that in our result, the appearance of b\sqrt{b} instead of bb forces us to reconcile the equation at the point x=y=π4x=y=\frac{\pi}{4}.

Substitute x=y=π/4x=y=\pi/4 into the original equation:

(a+2b⋅12)(a−2b⋅12)=(a+b)(a−b)=a2−b.\Bigl(a+\sqrt{2b}\cdot\frac{1}{\sqrt{2}}\Bigr) \Bigl(a-\sqrt{2b}\cdot\frac{1}{\sqrt{2}}\Bigr) =(a+\sqrt{b})(a-\sqrt{b})=a^2-b.

But it is given that the product equals a2−b2a^2-b^2. For these to be equal we must have:

a2−b=a2−b2⟹b=b2,a^2-b=a^2-b^2\quad\Longrightarrow\quad b=b^2,

which forces

b=1(since b>0).b=1 \quad (\text{since } b>0).

Then,

dxdy=a+1a−1=a+1a−1.\frac{dx}{dy}=\frac{a+\sqrt{1}}{a-\sqrt{1}}=\frac{a+1}{a-1}.

This matches option B because if we set b=1b=1 then

a+ba−b=a+1a−1.\frac{a+b}{a-b}=\frac{a+1}{a-1}.