The equation of normal at the point (1,1) on the curve 2y+x^3=3 is :

The correct answer is Option (3) → $3y - 2x=1$

The given curve is,

$2y+x^3=3$

Differentiating this w.r.t. 'x',

$⇒2\frac{dy}{dx}+3x^2=0$

$⇒\frac{dy}{dx}=-\frac{3x^2}{2}$

Now, the slope of normal to the curve is,

$\frac{dy}{dx}=\frac{2}{3x^2}$

$\left.\frac{dy}{dx}\right|_{(1,1)}=\frac{2}{3}$

Equation of the normal,

$y-y_1=m_{normal}(x-x_1)$

$y-1=\frac{2}{3}(x-1)$