we know that the equation of normal at $(am^2,-2am)$ is $y=mx-2am-am^3.$ If it passes through the point $(h,k)$ then.. $$am^3+0m^2+m(2a-h)+k=0 \tag1$$
above is a cubic equation in $m$ with three values of $m.$ Hence from point $(h,k)$ three normals can be drawn who slopes are given by [1].
also $m_1+m_2+m_3=\frac{0}{a}=0 $
what's the reason for the sum of the slopes of these normals is zero????