The slope of the normal to the curve $x^2+3y+y^2=5$ at (1, 1) is :

$\frac{5}{2}$

The correct answer is Option (1) → $\frac{5}{2}$

$x^2+3y+y^2=5$  P(1, 1)

differentiating wrt x

$2x+3\frac{dy}{dx}+2y\frac{dy}{dx}=0⇒\frac{dy}{dx}=\frac{-2x}{3+2y}$ slope of tangent

Slope of normal = $\left.-\frac{dy}{dx}=\frac{3+2y}{2x}\right]-{(1, 1)}=\frac{5}{2}$