Parabola from focus and directrix

Example — Parabola from the locus definition

Write the equation of the parabola with focus $F(0;0)$ and directrix $d:\ y = -4$, and verify that it corresponds to the expected form $y = ax^2 + bx + c$.

Locus condition: for a generic point $P(x;y)$, $d(P, F) = d(P, d) \implies \sqrt{x^2 + y^2} = \frac{|y + 4|}{\sqrt{0^2 + 1^2}} = |y+4|.$ Both sides are $\ge 0$ (the first is a distance, the second an absolute value): we may square directly. $x^2 + y^2 = (y+4)^2 = y^2 + 8y + 16 \implies x^2 = 8y + 16 \implies y = \tfrac{x^2}{8} - 2.$ It is indeed a parabola $y = ax^2 + bx + c$ with $a = \tfrac{1}{8}$, $b = 0$, $c = -2$.

Vertex: $V(0;-2)$, as one would expect geometrically (the point of the parabola closest to the directrix, halfway between $F$ and $d$).

The parabola $y = \tfrac{x^2}{8} - 2$ with focus $F$, vertex $V$ and directrix $d$: for every point $P$ the two distances $d(P,F)$ and $d(P,d)$ are equal.

Links

Topics: Geometric loci
Concepts: Directrix · Point-line distance · Focus · Parabola · Line · Vertex
Skills: Analytical geometry · Solving equations · Sketching graphs