Math Core

A satellite dish, a car headlight and a solar cooker all share one shape. Every one of them is a parabola, chosen for a single geometric property: signals arriving parallel to the axis bounce off the curve and all meet at one point, the focus. In this lesson you'll see the parabola defined by that focus, write its equation, and read its features straight from the equation.

A parabola from a point and a line

You already know parabolas as graphs of quadratic functions like $y = x^2$ . Geometry gives a second description that doesn't depend on any formula.

Definition

Parabola

A parabola is the set of all points in a plane that are the same distance from a fixed point, the focus, and a fixed line, the directrix. The focus does not lie on the directrix.

The point of the parabola halfway between the focus and the directrix is the vertex. The line through the focus perpendicular to the directrix is the axis of symmetry. The signed distance from the vertex to the focus is called $p$ . The directrix sits the same distance $|p|$ on the other side of the vertex.

Deriving the equation

Put the vertex at the origin, the focus at $(0, p)$ and the directrix on the line $y = -p$ . A point $(x, y)$ is on the parabola when its distance to the focus equals its distance to the directrix:

\sqrt{(x - 0)^2 + (y - p)^2} = |y + p|.

Square both sides and expand:

\begin{aligned} x^2 + y^2 - 2py + p^2 &= y^2 + 2py + p^2 \\ x^2 &= 4py. \end{aligned}

So $x^2 = 4py$ is a parabola with vertex $(0, 0)$ and focus $(0, p)$ . If $p > 0$ it opens up; if $p < 0$ it opens down. Swapping the roles of $x$ and $y$ gives $y^2 = 4px$ , which opens right ( $p > 0$ ) or left ( $p < 0$ ), with focus $(p, 0)$ and directrix $x = -p$ .

Shifting the vertex to $(h, k)$ replaces $x$ with $x - h$ and $y$ with $y - k$ .

Standard forms of a parabola with vertex (h, k)

Equation	Opens	Focus	Directrix	Axis
$(x - h)^2 = 4p(y - k)$	up if $p > 0$ , down if $p < 0$	$(h,\ k + p)$	$y = k - p$	$x = h$
$(y - k)^2 = 4p(x - h)$	right if $p > 0$ , left if $p < 0$	$(h + p,\ k)$	$x = h - p$	$y = k$

The squared variable tells you the axis: if $x$ is squared the parabola opens vertically, and if $y$ is squared it opens horizontally.

The chord through the focus perpendicular to the axis is the latus rectum. Its length is $|4p|$ , so its endpoints are $|2p|$ units on each side of the focus. These two points give you a quick way to sketch the width of the curve.

Worked example: Reading a parabola centered at the origin

Find the focus and directrix of $x^2 = 12y$ , then sketch it.

Solution. The equation has the form $x^2 = 4py$ with $4p = 12$ , so $p = 3$ . Since $x$ is squared and $p > 0$ , the parabola opens up.

Vertex: $(0, 0)$
Focus: $(0, 3)$
Directrix: $y = -3$

The latus rectum has length $12$ , so its endpoints are $6$ units to each side of the focus: $(-6, 3)$ and $(6, 3)$ . Check one: $6^2 = 36 = 12 \cdot 3$ . ✓

x² = 12y with its focus, vertex and the endpoints of the latus rectum. The directrix is dashed.Open in grapher →

Worked example: A shifted parabola that opens sideways

Find the vertex, focus and directrix of $(y + 1)^2 = -8(x - 2)$ .

Solution. Match the form $(y - k)^2 = 4p(x - h)$ . Here $k = -1$ , $h = 2$ and $4p = -8$ , so $p = -2$ .

Because $y$ is squared, the parabola opens horizontally. Because $p < 0$ , it opens left.

Vertex: $(2, -1)$
Focus: $(h + p, k) = (2 - 2, -1) = (0, -1)$
Directrix: $x = h - p = 2 - (-2)$ , so $x = 4$

The focus is inside the curve and the directrix is behind the vertex, on the side the parabola opens away from.

(y + 1)² = −8(x − 2) opens left. The directrix x = 4 is dashed.Open in grapher →

Common mistake

The number in front is $4p$ , not $p$ . In $(y + 1)^2 = -8(x - 2)$ , the focus is $2$ units from the vertex, not $8$ . Always divide by $4$ before you locate the focus and directrix.

Writing the equation from the focus and directrix

To go the other way, use the facts that the vertex is the midpoint between the focus and the directrix, and $p$ is the signed distance from the vertex to the focus.

Worked example: Build the equation

Write an equation of the parabola with focus $(3, 5)$ and directrix $y = 1$ .

Solution. The directrix is horizontal, so the parabola opens vertically and has the form $(x - h)^2 = 4p(y - k)$ .

The vertex lies halfway between the focus and the directrix, directly below the focus: $h = 3$ and $k = \dfrac{5 + 1}{2} = 3$ . The focus is $2$ units above the vertex, so $p = 2$ and $4p = 8$ :

(x - 3)^2 = 8(y - 3).

Check with the vertex: its distance to the focus is $5 - 3 = 2$ , and its distance to $y = 1$ is $3 - 1 = 2$ . ✓

From general form to standard form

Parabolas often show up expanded, like $y^2 - 4y - 8x + 28 = 0$ . Only one variable is squared, so this is a parabola. To find its features, complete the square in the squared variable and factor the other side.

Worked example: Complete the square

Find the vertex, focus and directrix of $y^2 - 4y - 8x + 28 = 0$ .

Solution. Keep the $y$ -terms on the left and move the rest to the right:

y^2 - 4y = 8x - 28.

Half of $-4$ is $-2$ , and $(-2)^2 = 4$ . Add $4$ to both sides:

\begin{aligned} y^2 - 4y + 4 &= 8x - 24 \\ (y - 2)^2 &= 8(x - 3). \end{aligned}

Now $h = 3$ , $k = 2$ and $4p = 8$ , so $p = 2$ . The parabola opens right.

Vertex: $(3, 2)$
Focus: $(5, 2)$
Directrix: $x = 1$

Tip

After completing the square, always factor out the coefficient on the other side. The form $(y - 2)^2 = 8x - 24$ hides the vertex; $(y - 2)^2 = 8(x - 3)$ shows it.

The reflective property

Every ray that travels parallel to the axis of a parabola reflects off the curve and passes through the focus. Run it backward and every ray leaving the focus reflects out parallel to the axis. That's why a dish antenna puts its receiver at the focus, and a flashlight puts its bulb there.

To place the receiver, model a cross-section of the dish as $x^2 = 4py$ with the vertex at the bottom. A point on the rim gives you $p$ , and $p$ is the distance from the vertex to the receiver.

Practice

Practice 1

Find the focus of the parabola $x^2 = -20y$ .

Enter a point like (2, -3)

Practice 2

Which way does the parabola $(x + 1)^2 = -6(y - 4)$ open?

Practice 3

The directrix of $(y - 3)^2 = 12(x + 2)$ is the line $x = c$ . Find $c$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Which equation describes the parabola with focus $(-4, 1)$ and directrix $x = 2$ ?

Practice 5

Find the focus of the parabola $x^2 + 6x - 4y + 1 = 0$ .

Enter a point like (2, -3)

Practice 6

A parabola has vertex $(2, -3)$ , a horizontal axis of symmetry, and passes through the point $(4, 1)$ . Find its focus.

Enter a point like (2, -3)

Practice 7

A satellite dish has a parabolic cross-section. It is $30$ cm across at the rim and $5$ cm deep at the center. How far from the vertex should the receiver be placed, in centimeters?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.