If a+2 bcosxa−2 bcosy=a2−b2, where a>b>0, then dxdy at π4, π4 is

a+2bcosxa−2bcosy=a2−b2−2bsinxa−2bcosy+a+2bcosx2bsinydydx=0At π4, π4, −ba−b+a+bb.dydx=0⇒bb−adxdy=−ba+b⇒dxdy=a+ba−b.

If a+2 bcosxa−2 bcosy=a2−b2, where a>b>0, then dxdy at π4, π4 is

a+2bcosxa−2bcosy=a2−b2−2bsinxa−2bcosy+a+2bcosx2bsinydydx=0At π4, π4, −ba−b+a+bb.dydx=0⇒bb−adxdy=−ba+b⇒dxdy=a+ba−b.

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If $(a + \sqrt{2} bcosx) (a - \sqrt{2} bcosy) = a^{2} - b^{2},$ where $a > b > 0,$ then $\frac{dx}{dy}$ at $(\frac{π}{4}, \frac{π}{4})$ is

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$\frac{2 a + b}{2 a - b}$

$\frac{a - 2 b}{a + 2 b}$

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

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

answer is B.

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Detailed Solution

$\begin{array}{l} (a + \sqrt{2} bcosx) (a - \sqrt{2} bcosy) = a^{2} - b^{2} \\ (- \sqrt{2} bsinx) (a - \sqrt{2} bcosy) + (a + \sqrt{2} bcosx) (\sqrt{2} bsiny) \frac{dy}{dx} = 0 \\ At (\frac{π}{4}, \frac{π}{4}), (- b) (a - b) + (a + b) b . \frac{dy}{dx} = 0 \\ \Rightarrow b (b - a) \frac{dx}{dy} = - b (a + b) \\ \Rightarrow \frac{dx}{dy} = \frac{a + b}{a - b} . \end{array}$