**Focus and directrix**
The ellipse and the hyperbola are often defined using two points, each of which is called a focus. The combined distances from these foci is used to create an equation of the ellipse and hyperbola.

A parabola has one focus point. The graph wraps around this focus. The equation of a parabola can be created using a combination of distances from the focus and from a line, called the directrix, to the graph. Examples are shown below, defining a parabola and creating its equation in this manner.

**Why we do this:**
One application of a parabola involves lighting. The focus of a parabola has a property which is important is constructing lamps and lighting. An automobile headlight is constructed of a reflective surface which is parabolic. The actual light bulb is placed at the focus. The light from this bulb reflects off the surface in parallel beams which “focus” the headlight beam directly forward from the car. This is because any light source at the focus of a parabola reflects in beams parallel to the axis of symmetry of the parabola.

A

parabola can be defined as the set of all points such that the distance from a

point on the

parabola to a focus

point is the same as the distance from the same

point on the

parabola to a fixed

line called the directrix. We illustrate this using a focus at the

point (0, 1) and a directrix given by the

equation y = –1.

On the

graph shown below the distances to the

point (x, y) from the focus and the directrix

line (the dotted line) must be equal. We will now use this information to derive an

equation of this parabola.

**Derivation:** Since the focus is (0, 1) and a general point on the directrix can be given by (x, -1), the distance formula would give us . Expanding this expression yields .

Simplifying by subtracting the common terms y^{2}, 1, as well as -2y from both sides gives x^{2} = 4y.

Solving this equation for y will produce this equation for our parabola where in this case, .

Examination of the equation tells us that the vertex of this parabola is the origin. The focal distance [from the vertex (0,0) to the focus (0,1)] is 1. Note that the equation could be rewritten as where d is the focal distance of 1.

This general form of the

equation for a

parabola tells us that if

, then the focal distance d is

because

. In this case, the focus is located at

and the

equation of the directrix is

. The

vertex is at the origin, (0, 0). A

graph of this quadratic is shown below.

**Let's Practice:** We will now examine two examples relating the equation, vertex, and focal distance.

- Let . Find the vertex, the focus, and the directrix.
**Step 1:** Find the vertex by completing the square. This equation can be rewritten as . The vertex is clearly (-1, -5).

**Step 2:** Solve for the focal length using the fact that .

**Step 3:** Since the graph of the parabola opens upward from the vertex, the focus is located at which is above the vertex.

**Step 4:** The directrix is located below the vertex at . Thus the equation of the directrix is .

- Suppose a vertex is located at (3, 1) and the focus is located at (3, 3). Find the directrix and an equation for this parabola.
**Step 1:** The distance from the vertex to the focus is 2 = d, the focal distance. Thus the directrix is located 2 units in the opposite direction from the vertex at y = -1.

**Step 2:**Vertex form of the equation of a parabola is given by where (h, k) are the coordinates of the vertex. We have .

**Step 3:** We know that . Thus our equation is .