Question

If $(a+\sqrt{2}b\cos x)(a-\sqrt{2}b\cos y)=a^2-b^2$ where $a>b>0,$ then $\frac{dx}{dy}$ at $\left(\frac{\pi }{4},\frac{\pi }{4}\right)$ is :

• A. $\frac{a-b}{a+b}$
• B. $\frac{a+b}{a-b}$
• C. $\frac{2a+b}{2a-b}$
• D. $\frac{a-2b}{a+2b}$

Answer 1

Answer: (B)

$(a+\sqrt{2}b\cos x)(a-\sqrt{2}b\cos y)=a^2-b^2$
$\Rightarrow a^2-\sqrt{2}ab\cos y+\sqrt{2}ab\cos x$
$-2b^2\cos x\cos y=a^2-b^2$
Differentiating both sides :
$0-\sqrt{2}ab\left(-\sin y\frac{dy}{dx}\right)+\sqrt{2}ab(-\sin x)$
$-2b^2\left[\cos x\left(-\sin y\frac{dy}{dx}\right)+\cos y(-\sin x)\right]=0$
At $\left(\frac{\pi }{4},\frac{\pi }{4}\right):$
$ab\frac{dy}{dx}-ab-2b^2\left(-\frac{1}{2}\frac{dy}{dx}-\frac{1}{2}\right)=0$
$\Rightarrow \frac{dx}{dy}=\frac{ab+b^2}{ab-b^2}=\frac{a+b}{a-b};a,b>0$