DWG NO. 8.2 — Lesson 2 of 4

Parabolas

Unit 8: Conic Sections · ~25 min

Objective Write the standard (vertex) form equation of a parabola from its focus and directrix, and identify the vertex, focus, and directrix from a standard-form equation.

Unit 2 graphed y = ax² + bx + c and called the result a parabola, but that equation only tells you where the curve is — not why it has that shape. The geometric definition fills that gap: a parabola is the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix). That definition, unlike "y equals a quadratic in x," works just as well for parabolas that open sideways.

Standard forms

Every parabola has a vertex (h, k) exactly halfway between its focus and directrix, and a value p equal to the distance from the vertex to the focus (and also from the vertex to the directrix).

Axis	Standard form	Opens	Focus	Directrix
Vertical	(x−h)² = 4p(y−k)	up if p>0, down if p<0	(h, k+p)	y = k−p
Horizontal	(y−k)² = 4p(x−h)	right if p>0, left if p<0	(h+p, k)	x = h−p

The squared variable tells you the axis: if x is squared, the parabola opens up or down (it's a function, same as Unit 2); if y is squared, it opens left or right (and fails the vertical line test — not a function, but still a parabola).

Parabola x² = 4(1.5)y — vertex at origin, focus above at (0, 1.5), directrix y = −1.5 below

Worked Example 1 · Vertex, focus, directrix from an equation

ProblemFind the vertex, focus, and directrix of (x−2)² = 8(y+1).

1Vertex: (h, k) = (2, −1).

2Match 4p = 8 → p = 2. Since x is squared and p > 0, the parabola opens upward.

3Focus: (h, k+p) = (2, −1+2) = (2, 1).

4Directrix: y = k−p = −1−2 = −3.

Vertex (2,−1) · Focus (2,1) · Directrix y = −3

Worked Example 2 · Equation from focus and directrix

ProblemWrite the equation of the parabola with focus (3, 0) and directrix x = −3.

1The vertex sits halfway between the focus and directrix, on the axis: (0, 0).

2p is the distance from vertex to focus: p = 3. The focus is right of the directrix, so the parabola opens right, and y is the squared variable.

3(y−0)² = 4(3)(x−0).

y² = 12x

Guided practice

1.Find the vertex, focus, and directrix of (x+1)² = −12(y−2).
Show answer

Vertex (−1, 2). 4p = −12 → p = −3, so it opens downward. Focus: (−1, 2−3) = (−1, −1). Directrix: y = 2+3 = 5.
2.Find the vertex, focus, and directrix of y² = 16x.
Show answer

Vertex (0, 0). 4p = 16 → p = 4, opens right. Focus (4, 0). Directrix x = −4.
3.Write the equation of the parabola with focus (0, 5) and directrix y = −5.
Show answer

Vertex (0,0), p = 5, opens upward, x is squared: x² = 20y.
4.Write the equation of the parabola with vertex (2, 3) and focus (2, 5).
Show answer

Focus is directly above the vertex, so it opens upward with p = 5−3 = 2. (x−2)² = 8(y−3).
5.A satellite dish's cross-section is a parabola with vertex at the origin, opening upward, and its focus 2 ft above the vertex. Write its equation and find the directrix.
Show answer

p = 2: x² = 8y. Directrix: y = −2.

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