$(x^2+y^2)^2$ = xy differentiating w.r.t. x ⇒ 2 $(x^2+y^2)$ $\left(2x+2y \frac{dy}{dx}\right)$ = y + x $\frac{dy}{dx}$ ⇒ $4x^3+4x^2yy \frac{dy}{dx}$ + $4y^2x$ + $4y^3 \frac{dy}{dx}$ - y - x $\frac{dy}{dx}$ = 0 ⇒ $(4x^2 y+4y^3-x)$ $\frac{dy}{dx}$ + $4x^3 + 4y^2x$ - y = 0 ⇒ $(4x^2 y+4y^3-x)$ $\frac{dy}{dx}$ = - $4x^3 - 4y^2 x$ + y ⇒ $\frac{dy}{dx}$ = $\frac{-4x^3-4y^2 x+y}{4x^2 y+4y^3 - x}$ OR x = a (2θ - sin 2θ) y = a (1 - cos θ) differentiating w.r.t. θ $\frac{dx}{d\theta}$ = a (2 - cos 2θ × 2) $\frac{dy}{d\theta}$ = a (+ sin 2θ × 2) $\frac{dy}{dx}$ = $\frac{+\sin 2\theta \times 2}{-2\cos 2\theta \times 2}$ = $\frac{\sin 2\theta}{1-\cos 2\theta}$ $\frac{dy}{dx}$ = $\frac{2\sin\theta \times \cos\theta}{2\sin^2\theta}$ $\frac{dy}{dx}$ = $\frac{\cos\theta}{\sin\theta}$ = cot θ at , θ = $\frac{\pi}{3}$ $\frac{dy}{dx}$ = cot $\frac{\pi}{3}$ = $\frac{1}{\sqrt{3}}$