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Study Guides > College Algebra

Graphs of Ellipses

Learning Objectives

  • Sketch a graph of an ellipse centered at the origin
  • Sketch a graph of an ellipse not centered at the origin
  • Express the equation of an ellipse in standard form given the equation in general form
Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form x2a2+y2b2=1, a>b\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\text{ }a>b for horizontal ellipses and x2b2+y2a2=1, a>b\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,\text{ }a>b for vertical ellipses.

How To: Given the standard form of an equation for an ellipse centered at (0,0)\left(0,0\right), sketch the graph.

  • Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
    • If the equation is in the form x2a2+y2b2=1\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1, where a>ba>b, then
      • the major axis is the x-axis
      • the coordinates of the vertices are (±a,0)\left(\pm a,0\right)
      • the coordinates of the co-vertices are (0,±b)\left(0,\pm b\right)
      • the coordinates of the foci are (±c,0)\left(\pm c,0\right)
    • If the equation is in the form x2b2+y2a2=1\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1, where a>ba>b, then
      • the major axis is the y-axis
      • the coordinates of the vertices are (0,±a)\left(0,\pm a\right)
      • the coordinates of the co-vertices are (±b,0)\left(\pm b,0\right)
      • the coordinates of the foci are (0,±c)\left(0,\pm c\right)
  • Solve for cc using the equation c2=a2b2{c}^{2}={a}^{2}-{b}^{2}.
  • Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.

Example: Graphing an Ellipse Centered at the Origin

Graph the ellipse given by the equation, x29+y225=1\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1. Identify and label the center, vertices, co-vertices, and foci.

Answer: First, we determine the position of the major axis. Because 25>925>9, the major axis is on the y-axis. Therefore, the equation is in the form x2b2+y2a2=1\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1, where b2=9{b}^{2}=9 and a2=25{a}^{2}=25. It follows that:

  • the center of the ellipse is (0,0)\left(0,0\right)
  • the coordinates of the vertices are (0,±a)=(0,±25)=(0,±5)\left(0,\pm a\right)=\left(0,\pm \sqrt{25}\right)=\left(0,\pm 5\right)
  • the coordinates of the co-vertices are (±b,0)=(±9,0)=(±3,0)\left(\pm b,0\right)=\left(\pm \sqrt{9},0\right)=\left(\pm 3,0\right)
  • the coordinates of the foci are (0,±c)\left(0,\pm c\right), where c2=a2b2{c}^{2}={a}^{2}-{b}^{2} Solving for cc, we have:

c=±a2b2=±259=±16=±4\begin{array}{l}c=\pm \sqrt{{a}^{2}-{b}^{2}}\hfill \\ =\pm \sqrt{25 - 9}\hfill \\ =\pm \sqrt{16}\hfill \\ =\pm 4\hfill \end{array}

Therefore, the coordinates of the foci are (0,±4)\left(0,\pm 4\right). Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Try It

Graph the ellipse given by the equation x236+y24=1\frac{{x}^{2}}{36}+\frac{{y}^{2}}{4}=1. Identify and label the center, vertices, co-vertices, and foci.

Answer: center: (0,0)\left(0,0\right); vertices: (±6,0)\left(\pm 6,0\right); co-vertices: (0,±2)\left(0,\pm 2\right); foci: (±42,0)\left(\pm 4\sqrt{2},0\right)

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