Normal at Parametric Point t

Normal to y² = 4ax at (at², 2at). Slope of this normal is −t. The foot of the normal on the axis is at (2a + at², 0).

At the point $(at^2, 2at)$, the normal slope is $-y_1/2a = -2at/2a = -t$.

Normal through $(at^2, 2at)$ with slope $-t$:

$$y - 2at = -t(x - at^2)$$ $$y - 2at = -tx + at^3$$ $$y + tx = 2at + at^3$$

Where does this normal meet the parabola again?

Substitute the normal $y = 2at + at^3 - tx$ into $y^2 = 4ax$. After simplification, the normal at $t$ meets the parabola at parameter $t'$ satisfying:

$$t' = -t - \frac{2}{t}$$

This is the "other end" of the normal chord. Note $t' \neq t$ (the normal is not a focal chord in general).

Condition for normal to pass through a given point $(h, k)$:

Substituting $(h, k)$ into $y + tx = 2at + at^3$:

$$k + th = 2at + at^3$$ $$at^3 + t(2a - h) + k = 0$$

This cubic in $t$ can have one or three real roots — hence one or three normals from a given point.