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

The correct answer is Option (2) → $x-y=0$

$2 y+x^2=3$

differentiating w.r.t. x

$⇒\frac{2dy}{dx}+2x=3⇒\frac{dy}{dx}=-x$

⇒ Slope of normal = $\frac{1}{x}$

$\left.\frac{1}{x}\right]_{(1,1)}=1$

so equation of normal

$y-1=x-1$

so $x-y=0$