Find the number of distinct normals drawn from the point $\left(-2, 1\right)$ to the parabola $y^2-4x-2y-3=0$.

Given parabola is 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{y}^2-4x-2y-3=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{y}^2-2y=4x+3\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(y-1{)}^2=4(x+1)\end{array}$

The equation of any normal to the given parabola is

$\begin{array}{l}(y-1)=m(x+1)-2am-a{m}^3\\ \Rightarrow (y-1)=m(x+1)-2m-m^3\end{array}$

which is passing through $\left(-2, 1\right)$, so 

$\begin{array}{l}(1-1)=m(-2+1)-2m-m^3\\ \Rightarrow 0=-m-2m-m^3\\ \Rightarrow m^3+3m=0\end{array}$

Hence, the number of distinct normals is $3$.