Analytic Geometry

Parabolas as Conic Sections

Parts of a Parabola

A parabola is a conic section defined in terms of distance to a fixed point and a fixed line.

The graphs of quadratic functions are parabolas. However, a parabola is defined differently when it is considered as a conic section. A parabola as a conic section is the set of points such that the distance from a fixed point, the focus, is equal to the distance to a fixed line, the directrix.

The properties of a parabola include:

  • For any point on a parabola, the distance from the point to the focus is equal to the distance from that same point to the directrix.
  • The vertex of a parabola is the midpoint of the segment from the focus to the directrix.
  • The axis of symmetry of the parabola divides the parabola into two halves that are mirror images. It passes through both the focus and the vertex.
A parabola is a type of conic section. The distance from one point on the parabola is the same as the distance from that point to a fixed line, called the directrix. The axis of symmetry goes through both the focus and the vertex.
A parabola can have its vertex at the origin or elsewhere on the coordinate plane, and its axis of symmetry may be vertical or horizontal. The equation of a parabola can be written in terms of the focus and directrix. A parabola with a vertical axis of symmetry and its vertex at the origin has an equation that can be written in the form:
y=(14a)x2y=\left(\frac{1}{4a}\right)x^2
It can be rewritten as:
x2=4ayx^2=4ay
  • If a>0a>0, the parabola opens upward.
  • If a<0a<0, the parabola opens downward.
  • The focus is at (0,a)(0,a), and the directrix is at y=ay=-a.

Parabolas with a Vertical Axis of Symmetry

Opens Upward Opens Downward
x2=8yx^2=8y x2=8yx^2=-8y
Focus: (0,2)(0,2) Focus: (0,2)(0,-2)
Directrix: y=2y = -2 Directrix: y=2y=2

For the parabola x2=8yx^2=8y or x2=4(2)yx^2=4(2)y, the value of aa is 2. Since aa is positive, the parabola opens upward and its focus is above the vertex. For the parabola x2=8yx^2=-8y or x2=4(2)yx^2=4(-2)y, the value of aa is –2. Since aa is negative, the parabola opens downward, and its focus is below the vertex. Both parabolas have a vertical axis of symmetry and a horizontal directrix.

A parabola with a horizontal axis of symmetry and its vertex at the origin has an equation that can be written in the form:
x=(14a)y2x=\left(\frac{1}{4a}\right)y^2
It can also be rewritten as:
y2=4axy^2=4ax
  • If a>0a>0, the parabola opens to the right.
  • If a<0a<0, the parabola opens to the left.
  • The focus is at (a,0)(a,0), and the directrix is at x=ax=-a.

Parabolas with a Horizontal Axis of Symmetry

Opens Right Opens Left
y2=8xy^2=8x y2=8xy^2=-8x
Focus: (2,0)(2,0) Focus: (2,0)(-2,0)
Directrix: x=2x=-2 Directrix: x=2x=2

For the parabola y2=8xy^2=8x or y2=4(2)xy^2=4(2)x, the value of aa is 2. Since aa is positive, the parabola opens to the right, and its focus is to the right of the vertex. For the parabola y2=8xy^2=-8x or y2=4(2)xy^2=4(-2)x, the value of aa is –2. Since aa is negative, the parabola opens to the left, and its focus is to the left of the vertex. Both parabolas have a horizontal axis of symmetry and a vertical directrix.

Graphing Parabolas

A parabola with a given equation can be graphed in the coordinate plane by locating the vertex, focus, and directrix.
A parabola may or may not have its vertex at the origin. A parabola with a vertex at the point (h, k) represents a horizontal translation by h units and a vertical translation by k units of a parabola with its vertex at the origin.

Equations of Parabolas with Vertex at (h,k)(h,k)

Equation Axis of Symmetry Focus Directrix
(xh)2=4a(yk)(x-h)^2=4a(y-k)
Vertical: x=hx=h
(h,k+a)(h,k+a)
y=kay = k-a
(yk)2=4a(xh)(y-k)^2=4a(x-h)
Horizontal: y=ky=k
(h+a,k)(h+a,k)
x=hax = h-a

Step-By-Step Example
Graphing a Parabola
A parabola has the equation:
8(y2)=(x+5)28(y-2)=(x+5)^2
Calculate the vertex, focus, and directrix of the parabola. Sketch a graph of the parabola.
Step 1

Write the equation in standard form.

The squared variable in the equation is xx, so the parabola has a vertical axis of symmetry and the standard form of its equation is:
(xh)2=4a(yk)(x-h)^2=4a(y-k)
Rewrite the equation using the standard form:
8(y2)=(x+5)2(x+5)2=8(y2)(x(5))2=8(y2)(x(5))2=4(2)(y2)\begin{aligned}8(y-2)&=(x+5)^2\\(x+5)^2&=8(y-2)\\(x-(-5))^2&=8(y-2)\\(x-(-5))^2&=4(2)(y-2)\end{aligned}
Step 2
Use the standard form of the equation to identify the values of hh, kk, and aa.
(xh)2=4a(yk)(x(5))2=4(2)(y2)\begin{aligned}(x-h)^2&=4a(y-k)\\(x-(-5))^2&=4(2)(y-2)\end{aligned}
So, h=5h=-5, k=2k=2, and a=2a=2.
Step 3

Determine the vertex, focus, and directrix.

The vertex (h,k)(h, k) is (5,2)(-5, 2).

The coordinates of the focus are (h,k+a)(h, k+a).

The focus is (5,2+2)(-5, 2+2), or (5,4)(-5, 4).

The equation of the directrix is:
y=kay = k-a
The directrix is the horizontal line:
y=22=0y=2-2=0
Step 4

Plot the focus and vertex, and graph the directrix.

To locate another point on the parabola, draw a square with one vertex at the focus and one side along the directrix. The vertex (1,4)(-1, 4) of this square is a point on the parabola. Use symmetry to locate the point (9,4)(-9,4).
Solution
Draw the parabola through the three points.