Number of Normals from a Point

For the parabola $$y^2=4x,$$ we have $$a=1.$$

Let the foot of a normal be $$\left(t^2,2t\right).$$

The equation of the normal to $$y^2=4ax$$ is

$$y=-tx+2at+at^3$$

Since $$a=1,$$ the normal becomes

$$y=-tx+2t+t^3$$

Since the normal passes through $$P(3,0),$$ substitute $$x=3$$ and $$y=0.$$

Hence,

$$0=-3t+2t+t^3$$

$$t^3-t=0$$

$$t(t^2-1)=0$$

$$t(t-1)(t+1)=0$$

Thus,

$$t=0,1,-1$$

The ordinates of the feet of the normals are $$\alpha=2,$$ $$\beta=-2$$ and $$\gamma=0.$$

Therefore,

$$\alpha^2+\beta^2+\gamma^2=2^2+(-2)^2+0^2$$

$$=4+4+0$$

$$=8$$