GeometryCirclesG-GPE.A.2

Parabola from Focus and Directrix

A parabola is all points equally far from a point and a line.

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You graphed parabolas like $y = a(x - h)^2 + k$ in Vertex Form. Geometry gives them a definition in the same spirit as the circle in Equation of a Circle: a parabola is the set of points that are the same distance from a fixed point, the focus, and a fixed line, the directrix.

Why it works

Put the focus at $(0, p)$ and the directrix at $y = -p$. A point $(x, y)$ is on the parabola when $\sqrt{x^2 + (y - p)^2} = |y + p|$. Square both sides: $x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2$. The $y^2$ and $p^2$ cancel, leaving $x^2 = 4py$. The vertex is halfway between the focus and the directrix, at the origin. Shifting the vertex to $(h, k)$ gives $(x - h)^2 = 4p(y - k)$, which is vertex form with $a = \frac{1}{4p}$.
Move the sliders: the focus and directrix follow. A wider parabola (smaller $a$) has its focus farther from the vertex.

Key idea

A parabola is all points equally far from a focus and a directrix. With vertex $(h, k)$ and focus $p$ units above it (below if $p < 0$): $(x - h)^2 = 4p(y - k)$, and the directrix is $y = k - p$.

Worked example: Vertex at the origin

Write an equation of the parabola with focus $(0, 3)$ and directrix $y = -3$.
  1. The vertex is halfway between: $(0, 0)$. The focus is $p = 3$ above it.
  2. $x^2 = 4py = 12y$.

Answer: $x^2 = 12y$ (or $y = \frac{1}{12}x^2$)

Worked example: A shifted parabola

Write an equation of the parabola with focus $(2, 5)$ and directrix $y = 1$.
  1. Vertex: halfway between $y = 5$ and $y = 1$ at $x = 2$: $(2, 3)$. So $p = 5 - 3 = 2$.
  2. $(x - 2)^2 = 4 \cdot 2 (y - 3) = 8(y - 3)$.

Answer: $(x - 2)^2 = 8(y - 3)$

Worked example: Reading the focus and directrix

Find the focus and directrix of $(x + 1)^2 = -8(y - 4)$.
  1. Vertex $(-1, 4)$ and $4p = -8$, so $p = -2$: the parabola opens down.
  2. Focus: 2 units below the vertex, $(-1, 2)$. Directrix: 2 units above, $y = 6$.

Answer: Focus $(-1, 2)$; directrix $y = 6$

Common mistake

Forgetting that $p$ is the distance from the vertex to the focus, not from the focus to the directrix (that distance is $2p$). In the second example the focus and directrix are 4 apart, so $p = 2$. Check: any point on your parabola, such as the vertex, should be equally far from the focus and the directrix.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.

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Sources and further reading

  • OpenStax Intermediate Algebra 2e, 11.2 Parabolas
  • Common Core State Standards for Mathematics, High School Geometry, G-GPE.A.2
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