If (a+√2b x)(a-√2b y)=a^2-b^2, where ab0, then dxdy at (4,4) is

Question

If $(a+\sqrt2b\cos x)(a-\sqrt2b\cos y)=a^2-b^2$, where $a>b>0$, then $\dfrac{dx}{dy}$ at $\left(\dfrac\pi4,\dfrac\pi4\right)$ is

Accepted Answer

Differentiating: $-\sqrt2b\sin x(a-\sqrt2b\cos y)+(a+\sqrt2b\cos x)\sqrt2b\sin y\dfrac{dy}{dx}=0$.

At $\left(\dfrac\pi4,\dfrac\pi4\right)$: $\dfrac{dy}{dx}=\dfrac{a-b}{a+b}\Rightarrow\dfrac{dx}{dy}=\dfrac{a+b}{a-b}$.

Solutions · 1