The normal at the point $(1,\;1)$ on the curve $2y + {x^2} = 3$ is

(A) $x + y = 0$

(B) $x - y = 0$

(C) $x + y + 1 = 0$

(D) $- x + y + 2 = 0$

Option B is correct

We have, $2y + {x^2} = 3$ …(i)

Differentiating (i) w.r.t. $x$, we get $2\cfrac{{dy}}{{dx}} + 2x = 0 \Rightarrow \cfrac{{dy}}{{dx}} = - x$

As, ${\left( {\cfrac{{dy}}{{dx}}} \right)_{(1,1)}} = - 1$.

Therefore llope of normal $= 1$

The equation of normal is given by :

$y - 1 = 1 \cdot (x - 1) \Rightarrow x - y = 0.$