Foot of Perpendicular from Focus to Tangent

For y² = 4ax, the foot of the perpendicular drawn from the focus to any tangent always lies on the y-axis (the tangent at the vertex). Equivalently, the tangent at the vertex is the locus of feet of perpendiculars from the focus.

For $y^2 = 4ax$, the tangent at parameter $t$ is $ty = x + at^2$, or $x - ty + at^2 = 0$.

The focus is $S(a, 0)$. The foot of the perpendicular from $S$ to the tangent:

The perpendicular from $S(a, 0)$ to the line $x - ty + at^2 = 0$ has direction $(1, -t)$ (the normal to the tangent). Parametric form of the perpendicular:

$$x = a + \lambda, \quad y = 0 - t\lambda = -t\lambda$$

Substituting into the tangent equation:

$$(a + \lambda) - t(-t\lambda) + at^2 = 0$$ $$a + \lambda + t^2\lambda + at^2 = 0$$ $$a(1 + t^2) + \lambda(1 + t^2) = 0$$ $$\lambda = -a$$

Foot of perpendicular:

$$x = a + (-a) = 0, \quad y = -t(-a) = at$$

The foot is $(0, at)$ — which lies on the $y$-axis (the tangent at the vertex) for every value of $t$.

Conclusion: As the tangent varies over all points of $y^2 = 4ax$, the foot of the perpendicular from the focus traces the $y$-axis. The $y$-axis is therefore the locus of feet of perpendiculars from the focus to all tangents.