Formulae for focus, vertex and directrix

Property — General formulae

Given the parabola $y = ax^2 + bx + c$, one has: $V\left(-\frac{b}{2a};\ -\frac{\Delta}{4a}\right),\qquad F\left(-\frac{b}{2a};\ -\frac{\Delta}{4a} + \frac{1}{4a}\right),\qquad d:\ y = -\frac{\Delta}{4a} - \frac{1}{4a}.$ The focus lies on the axis of symmetry, “inside” the curve, at distance $\tfrac{1}{|4a|}$ from the vertex; the directrix is perpendicular to the axis of symmetry, at the same distance but on the opposite side.

Proof

We start from the form $y = ax^2 + bx + c$ and rewrite it in “canonical form” by completing the square: $y = a\bigl(x - x_V\bigr)^2 + y_V$ with $x_V = -\tfrac{b}{2a}$ and $y_V = -\tfrac{\Delta}{4a}$. Translating the origin to the vertex we obtain $Y = aX^2$ and, from the locus definition, we verify that the focus has $Y_F = \tfrac{1}{4a}$ and the directrix $Y_d = -\tfrac{1}{4a}$. Returning to the original coordinates (translating back) we obtain the stated formulae. ∎

Links

Topics: Geometric loci
Concepts: Directrix · Focus · Parabola · Vertex
Skills: Proving · Analytical geometry · Using formulae