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Study Guides > College Algebra

Parabolas with Vertices at the Origin

Learning Objectives

  • Identify and label the focus, directrix, and endpoints of the focal diameter of a parabola
  • Write the equation of a parabola given a focus and directrix
In The Ellipse, we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola.
Parabola
Like the ellipse and hyperbola, the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points (x,y)\left(x,y\right) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix. We previously learned about a parabola’s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus. The line segment that passes through the focus and is parallel to the directrix is called the latus rectum, also called the focal diameter. The endpoints of the focal diameter lie on the curve. By definition, the distance dd from the focus to any point PP on the parabola is equal to the distance from PP to the directrix.
Key features of the parabola
To work with parabolas in the coordinate plane, we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former. Let (x,y)\left(x,y\right) be a point on the parabola with vertex (0,0)\left(0,0\right), focus (0,p)\left(0,p\right), and directrix y=−py= -p as shown in Figure 4. The distance dd from point (x,y)\left(x,y\right) to point (x,−p)\left(x,-p\right) on the directrix is the difference of the y-values: d=y+pd=y+p. The distance from the focus (0,p)\left(0,p\right) to the point (x,y)\left(x,y\right) is also equal to dd and can be expressed using the distance formula.

d=(x−0)2+(y−p)2=x2+(y−p)2\begin{array}{l}d=\sqrt{{\left(x - 0\right)}^{2}+{\left(y-p\right)}^{2}}\hfill \\ =\sqrt{{x}^{2}+{\left(y-p\right)}^{2}}\hfill \end{array}

Set the two expressions for dd equal to each other and solve for yy to derive the equation of the parabola. We do this because the distance from (x,y)\left(x,y\right) to (0,p)\left(0,p\right) equals the distance from (x,y)\left(x,y\right) to (x,−p)\left(x, -p\right). x2+(y−p)2=y+p\sqrt{{x}^{2}+{\left(y-p\right)}^{2}}=y+p We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.

x2+(y−p)2=(y+p)2x2+y2−2py+p2=y2+2py+p2x2−2py=2py x2=4py\begin{array}{c}{x}^{2}+{\left(y-p\right)}^{2}={\left(y+p\right)}^{2}\\ {x}^{2}+{y}^{2}-2py+{p}^{2}={y}^{2}+2py+{p}^{2}\\ {x}^{2}-2py=2py\\ \text{ }{x}^{2}=4py\end{array}

The equations of parabolas with vertex (0,0)\left(0,0\right) are y2=4px{y}^{2}=4px when the x-axis is the axis of symmetry and x2=4py{x}^{2}=4py when the y-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.

A General Note: Standard Forms of Parabolas with Vertex (0, 0)

The table below summarizes the standard features of parabolas with a vertex at the origin.
Axis of Symmetry Equation Focus Directrix Endpoints of Focal Diameter
x-axis y2=4px{y}^{2}=4px (p, 0)\left(p,\text{ }0\right) x=−px=-p (p, ±2p)\left(p,\text{ }\pm 2p\right)
y-axis x2=4py{x}^{2}=4py (0, p)\left(0,\text{ }p\right) y=−py=-p (±2p, p)\left(\pm 2p,\text{ }p\right)
(a) When p>0p>0 and the axis of symmetry is the x-axis, the parabola opens right. (b) When p<0p<0 and the axis of symmetry is the x-axis, the parabola opens left. (c) When p<0p<0 and the axis of symmetry is the y-axis, the parabola opens up. (d) When  p<0 \text{ }p<0\text{ } and the axis of symmetry is the y-axis, the parabola opens down.
The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and focal diameter. When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola. A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the focal diameter, these lines intersect on the axis of symmetry.

How To: Given a standard form equation for a parabola centered at (0, 0), sketch the graph.

  • Determine which of the standard forms applies to the given equation: y2=4px{y}^{2}=4px or x2=4py{x}^{2}=4py.
  • Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the focal diameter.
    • If the equation is in the form y2=4px{y}^{2}=4px, then
      • the axis of symmetry is the x-axis, y=0y=0
      • set 4p4p equal to the coefficient of x in the given equation to solve for pp. If p>0p>0, the parabola opens right. If p<0p<0, the parabola opens left.
      • use pp to find the coordinates of the focus, (p,0)\left(p,0\right)
      • use pp to find the equation of the directrix, x=−px=-p
      • use pp to find the endpoints of the focal diameter, (p,±2p)\left(p,\pm 2p\right). Alternately, substitute x=px=p into the original equation.
    • If the equation is in the form x2=4py{x}^{2}=4py, then
      • the axis of symmetry is the y-axis, x=0x=0
      • set 4p4p equal to the coefficient of y in the given equation to solve for pp. If p>0p>0, the parabola opens up. If p<0p<0, the parabola opens down.
      • use pp to find the coordinates of the focus, (0,p)\left(0,p\right)
      • use pp to find equation of the directrix, y=−py=-p
      • use pp to find the endpoints of the focal diameter, (±2p,p)\left(\pm 2p,p\right)
  • Plot the focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.

Example: Graphing a Parabola with Vertex (0, 0) and the x-axis as the Axis of Symmetry

Graph y2=24x{y}^{2}=24x. Identify and label the focus, directrix, and endpoints of the focal diameter.

Answer: The standard form that applies to the given equation is y2=4px{y}^{2}=4px. Thus, the axis of symmetry is the x-axis. It follows that:

  • 24=4p24=4p, so p=6p=6. Since p>0p>0, the parabola opens right the coordinates of the focus are (p,0)=(6,0)\left(p,0\right)=\left(6,0\right)
  • the equation of the directrix is x=−p=−6x=-p=-6
  • the endpoints of the focal diameter have the same x-coordinate at the focus. To find the endpoints, substitute x=6x=6 into the original equation: (6,±12)\left(6,\pm 12\right)
Next we plot the focus, directrix, and focal diameter, and draw a smooth curve to form the parabola.

Try It

Graph y2=−16x{y}^{2}=-16x. Identify and label the focus, directrix, and endpoints of the focal diameter.

Answer: Focus: (−4,0)\left(-4,0\right); Directrix: x=4x=4; Endpoints of the latus rectum: (−4,±8)\left(-4,\pm 8\right)

 

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