Graph Hyperbolas

Learning Outcomes

  • Graph a hyperbola centered at the origin.
  • Graph a hyperbola not centered at the origin.

When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex] for horizontal hyperbolas and the standard form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex] for vertical hyperbolas.

How To: Given a standard form equation for a hyperbola centered at [latex]\left(0,0\right)[/latex], sketch the graph.

  • Determine which of the standard forms applies to the given equation.
  • Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes.
    • If the equation is in the form [latex]\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1[/latex], then
      • the transverse axis is on the x-axis
      • the coordinates of the vertices are [latex]\left(\pm a,0\right)[/latex]
      • the coordinates of the co-vertices are [latex]\left(0,\pm b\right)[/latex]
      • the coordinates of the foci are [latex]\left(\pm c,0\right)[/latex]
      • the equations of the asymptotes are [latex]y=\pm \frac{b}{a}x[/latex]
    • If the equation is in the form [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex], then
      • the transverse axis is on the y-axis
      • the coordinates of the vertices are [latex]\left(0,\pm a\right)[/latex]
      • the coordinates of the co-vertices are [latex]\left(\pm b,0\right)[/latex]
      • the coordinates of the foci are [latex]\left(0,\pm c\right)[/latex]
      • the equations of the asymptotes are [latex]y=\pm \frac{a}{b}x[/latex]
  • Solve for the coordinates of the foci using the equation [latex]c=\pm \sqrt{{a}^{2}+{b}^{2}}[/latex].
  • Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola.

Example: Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form

Graph the hyperbola given by the equation [latex]\dfrac{{y}^{2}}{64}-\dfrac{{x}^{2}}{36}=1[/latex]. Identify and label the vertices, co-vertices, foci, and asymptotes.

Try It

Graph the hyperbola given by the equation [latex]\dfrac{{x}^{2}}{144}-\dfrac{{y}^{2}}{81}=1[/latex]. Identify and label the vertices, co-vertices, foci, and asymptotes.

Try It

Use an online graphing tool to plot the equation [latex]\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1[/latex]. Adjust the values you use for [latex]a,b[/latex] to values between [latex]1, 20[/latex]. Your task in this exercise is to graph a hyperbola and then calculate and add the following features to the graph:

  • vertices
  • co-vertices
  • foci
  • asymptotes

Graphing Hyperbolas Not Centered at the Origin

Graphing hyperbolas centered at a point [latex]\left(h,k\right)[/latex] other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms [latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex] for horizontal hyperbolas, and [latex]\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}-\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex] for vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes.

tip for success

Writing the equation of a hyperbola not centered at the origin uses graph transformation techniques in the same way that writing the equation of an ellipse does. The center is represented by [latex]\left(h, k\right)[/latex] as it was in the ellipse.

Look for similarities and differences in the construction of ellipses, hyperbolas, and parabolas to help you build your intuition about these objects.

How To: Given a general form for a hyperbola centered at [latex]\left(h,k\right)[/latex], sketch the graph.

  • Convert the general form to that standard form. Determine which of the standard forms applies to the given equation.
  • Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes.
    • If the equation is in the form [latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex], then
      • the transverse axis is parallel to the x-axis
      • the center is [latex]\left(h,k\right)[/latex]
      • the coordinates of the vertices are [latex]\left(h\pm a,k\right)[/latex]
      • the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)[/latex]
      • the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex]
      • the equations of the asymptotes are [latex]y=\pm \frac{b}{a}\left(x-h\right)+k[/latex]
    • If the equation is in the form [latex]\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}-\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}=1[/latex], then
      • the transverse axis is parallel to the y-axis
      • the center is [latex]\left(h,k\right)[/latex]
      • the coordinates of the vertices are [latex]\left(h,k\pm a\right)[/latex]
      • the coordinates of the co-vertices are [latex]\left(h\pm b,k\right)[/latex]
      • the coordinates of the foci are [latex]\left(h,k\pm c\right)[/latex]
      • the equations of the asymptotes are [latex]y=\pm \frac{a}{b}\left(x-h\right)+k[/latex]
  • Solve for the coordinates of the foci using the equation [latex]c=\pm \sqrt{{a}^{2}+{b}^{2}}[/latex].
  • Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola.

recall Complete the square

You will need to complete the square twice when rewriting the general form of a hyperbola in the same way you did with the ellipse. Here’s the form to help you.

Completing the Square

Given an expression of the form [latex]a\left(x^2+bx\right)[/latex], add [latex]\left(\dfrac{b}{2}\right)^2[/latex] inside the parentheses, then subtract [latex]a\left(\dfrac{b}{2}\right)^2[/latex] to counteract the change you made.

If completing the square on one side of an equation, you may either subtract the value of [latex]a\left(\dfrac{b}{2}\right)^2[/latex] from that side, or add it to the other to maintain equality.

Then factor the perfect square trinomial you created inside the original parentheses.

Example

[latex]\qquad a\left(x^2+bx\right)[/latex]

[latex]=a\left(x^2+bx+ \left(\dfrac{b}{2}\right)^2\right)-a\left(\dfrac{b}{2}\right)^2[/latex]

[latex]=a\left(x+ \dfrac{b}{2}\right)^2-a\left(\dfrac{b}{2}\right)^2[/latex]

Example: Graphing a Hyperbola Centered at (h, k) Given an Equation in General Form

Graph the hyperbola given by the equation [latex]9{x}^{2}-4{y}^{2}-36x - 40y - 388=0[/latex]. Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Try It

Graph the hyperbola given by the standard form of an equation [latex]\dfrac{{\left(y+4\right)}^{2}}{100}-\dfrac{{\left(x - 3\right)}^{2}}{64}=1[/latex]. Identify and label the center, vertices, co-vertices, foci, and asymptotes.

Try It

Using an online graphing calculator, plot a hyperbola not centered at the origin.

The equation used to generate the graph is [latex]\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1[/latex].

Adjust the values you use for [latex]a,b[/latex] to values between [latex]1, 20[/latex], and the variables [latex]h,k[/latex] to numbers between [latex]10,10[/latex]. Your task in this exercise is to graph the hyperbola and then calculate and add the following features to the graph:

  • center
  • vertices
  • co-vertices
  • foci
  • asymptotes