Conic Sections: Focus and Directrix

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. (LINK to ellipse and hyperbola lessons)

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 LINK would give us .

Expanding this expression yields .

Simplifying by subtracting the common terms y², 1, as well as -2y from both sides gives

x² = 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.

We will now examine two other examples relating the equation, vertex, and focal distance.

1. Let . Find the vertex, the focus, and the directrix.

   Step 1: Find the vertex (LINK) by completing the square (LINK). 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 (LINK) 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 .

2. 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 (LINK to Vertex of a Parabola) 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 .