If $P\left(-3,2\right)$ is one end of focal chord PQ of the parabola $y^2+4x+4y=0$ ,then the slope of normal at Q is

$$\begin{gathered} y^2+4x+4y=0 \\ differentiating wrt x \\ 2y\frac{ dy}{dx}+4+4\frac{ dy}{dx}=0\Rightarrow \frac{ dy}{dx}=\frac{-4}{2y+4}=\frac{-2}{y+2} \end{gathered}$$

Slope of normal at Q= slope of tangent at P $={\left(\frac{dy}{dx}\right)}_{\left(-3,2\right)}=\boxed{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$