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

# If $\left(\mathrm{a}+\sqrt{2}\text{\hspace{0.17em}}\mathrm{bcosx}\right)\left(\mathrm{a}-\sqrt{2}\text{\hspace{0.17em}}\mathrm{bcosy}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2},$ where $\mathrm{a}>\mathrm{b}>0,$ then $\frac{\mathrm{dx}}{\mathrm{dy}}$ at $\left(\frac{\mathrm{\pi }}{4},\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{\pi }}{4}\right)$ is

1. A

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

2. B

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

3. C

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

4. D

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

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### Solution:

$\begin{array}{l}\left(\mathrm{a}+\sqrt{2}\mathrm{bcosx}\right)\left(\mathrm{a}-\sqrt{2}\mathrm{bcosy}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\\ \left(-\sqrt{2}\mathrm{bsinx}\right)\left(\mathrm{a}-\sqrt{2}\mathrm{bcosy}\right)+\left(\mathrm{a}+\sqrt{2}\mathrm{bcosx}\right)\left(\sqrt{2}\mathrm{bsiny}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ \mathrm{At}\text{\hspace{0.17em}\hspace{0.17em}}\left(\frac{\mathrm{\pi }}{4},\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{\pi }}{4}\right),\text{\hspace{0.17em}\hspace{0.17em}}\left(-\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)+\left(\mathrm{a}+\mathrm{b}\right)\mathrm{b}.\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ ⇒\mathrm{b}\left(\mathrm{b}-\mathrm{a}\right)\frac{\mathrm{dx}}{\mathrm{dy}}=-\mathrm{b}\left(\mathrm{a}+\mathrm{b}\right)\\ ⇒\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}-\mathrm{b}}.\end{array}$  +91

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