Graph Hyperbolas

Learning Objectives

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]\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1[/latex] for horizontal hyperbolas and the standard form [latex]\frac{{y}^{2}}{{a}^{2}}-\frac{{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]\frac{{x}^{2}}{{a}^{2}}-\frac{{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]\frac{{y}^{2}}{{a}^{2}}-\frac{{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]\frac{{y}^{2}}{64}-\frac{{x}^{2}}{36}=1[/latex]. Identify and label the vertices, co-vertices, foci, and asymptotes.

Solution

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

Solution

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In the graph below you will find the plot of a Hyperbola centered at the origin. Note the equation used to generate the graph is [latex]\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1[/latex], and the variables [latex]a,b[/latex] are free to slide between [latex] 1, 20[/latex]. Your task in this exercise is to use the free variables to add the following features to the graph:

vertices
co-vertices
foci
asymptotes

For example, to plot the vertices, you would define two points like this: [latex](a,0)[/latex] and [latex](-a,0)[/latex] on their own lines.

https://www.desmos.com/calculator/du6j3vvisl

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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]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex] for horizontal hyperbolas, and [latex]\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\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.

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]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\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]\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\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.

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.

Solution

Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation.

[latex]\left(9{x}^{2}-36x\right)-\left(4{y}^{2}+40y\right)=388[/latex]

Factor the leading coefficient of each expression.

[latex]9\left({x}^{2}-4x\right)-4\left({y}^{2}+10y\right)=388[/latex]

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

[latex]9\left({x}^{2}-4x+4\right)-4\left({y}^{2}+10y+25\right)=388+36 - 100[/latex]

Rewrite as perfect squares.

[latex]9{\left(x - 2\right)}^{2}-4{\left(y+5\right)}^{2}=324[/latex]

Divide both sides by the constant term to place the equation in standard form.

[latex]\frac{{\left(x - 2\right)}^{2}}{36}-\frac{{\left(y+5\right)}^{2}}{81}=1[/latex]

The standard form that applies to the given equation is [latex]\frac{{\left(x-h\right)}^{2}}{{a}^{2}}-\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex], where [latex]{a}^{2}=36[/latex] and [latex]{b}^{2}=81[/latex], or [latex]a=6[/latex] and [latex]b=9[/latex]. Thus, the transverse axis is parallel to the x-axis. It follows that:

the center of the ellipse is [latex]\left(h,k\right)=\left(2,-5\right)[/latex]
the coordinates of the vertices are [latex]\left(h\pm a,k\right)=\left(2\pm 6,-5\right)[/latex], or [latex]\left(-4,-5\right)[/latex] and [latex]\left(8,-5\right)[/latex]
the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)=\left(2,-5\pm 9\right)[/latex], or [latex]\left(2,-14\right)[/latex] and [latex]\left(2,4\right)[/latex]
the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex], where [latex]c=\pm \sqrt{{a}^{2}+{b}^{2}}[/latex]. Solving for [latex]c[/latex], we have

[latex]c=\pm \sqrt{36+81}=\pm \sqrt{117}=\pm 3\sqrt{13}[/latex]

Therefore, the coordinates of the foci are [latex]\left(2 - 3\sqrt{13},-5\right)[/latex] and [latex]\left(2+3\sqrt{13},-5\right)[/latex].

The equations of the asymptotes are [latex]y=\pm \frac{b}{a}\left(x-h\right)+k=\pm \frac{3}{2}\left(x - 2\right)-5[/latex].

Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola.

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

Solution

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In the graph below you will find the plot of a Hyperbola not centered at the origin. Note the equation used to generate the graph is [latex]\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1[/latex], and the variables [latex]a,b[/latex] are free to slide between [latex] 1, 20[/latex], and the variables [latex]h,k[/latex] are free to slide between [latex]10,10[/latex]. Your task in this exercise is to use the free variables to add the following features to the graph:

center
vertices
co-vertices
foci
asymptotes

For example, to plot the center, you would define a point like this: [latex](h,k)[/latex].

https://www.desmos.com/calculator/r307oiry1a

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Conic Sections

Learning Objectives

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

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

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Graphing Hyperbolas Not Centered at the Origin

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

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

Try It

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