Q.Find the equations of tangent and normal to the curve at the point on it: $x\sin 2y = y\cos 2x$ at $\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$.

Given $x\sin 2y = y\cos 2x$. Differentiating both sides w.r.t. $x$:

LHS: $\sin 2y + x\cdot 2\cos 2y\,\dfrac{dy}{dx}$. RHS: $\dfrac{dy}{dx}\cos 2x - 2y\sin 2x$.

So $\sin 2y + 2x\cos 2y\,\dfrac{dy}{dx} = \dfrac{dy}{dx}\cos 2x - 2y\sin 2x$, giving $\dfrac{dy}{dx}(2x\cos 2y - \cos 2x) = -2y\sin 2x - \sin 2y$, so $\dfrac{dy}{dx} = \dfrac{-2y\sin 2x-\sin 2y}{2x\cos 2y-\cos 2x}$.

At $\left(\tfrac{\pi}{4},\tfrac{\pi}{2}\right)$: $2x=\tfrac{\pi}{2}$, $2y=\pi$. So $\sin 2x = \sin\tfrac{\pi}{2}=1$, $\cos 2y=\cos\pi=-1$, $\cos 2x=\cos\tfrac{\pi}{2}=0$, $\sin 2y=\sin\pi=0$.

Numerator $= -2\left(\tfrac{\pi}{2}\right)(1) - 0 = -\pi$. Denominator $= 2\left(\tfrac{\pi}{4}\right)(-1) - 0 = -\tfrac{\pi}{2}$. So $m = \dfrac{-\pi}{-\pi/2} = 2$. …

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