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 $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$ for horizontal hyperbolas and the standard form $\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1$ for vertical hyperbolas.

How To: Given a standard form equation for a hyperbola centered at $\left(0,0\right)$, 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 $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$, then
• the transverse axis is on the x-axis
• the coordinates of the vertices are $\left(\pm a,0\right)$
• the coordinates of the co-vertices are $\left(0,\pm b\right)$
• the coordinates of the foci are $\left(\pm c,0\right)$
• the equations of the asymptotes are $y=\pm \frac{b}{a}x$
• If the equation is in the form $\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1$, then
• the transverse axis is on the y-axis
• the coordinates of the vertices are $\left(0,\pm a\right)$
• the coordinates of the co-vertices are $\left(\pm b,0\right)$
• the coordinates of the foci are $\left(0,\pm c\right)$
• the equations of the asymptotes are $y=\pm \frac{a}{b}x$
• Solve for the coordinates of the foci using the equation $c=\pm \sqrt{{a}^{2}+{b}^{2}}$.
• 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 $\dfrac{{y}^{2}}{64}-\dfrac{{x}^{2}}{36}=1$. Identify and label the vertices, co-vertices, foci, and asymptotes.

Try It

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

https://ohm.lumenlearning.com/multiembedq.php?id=65294&theme=oea&iframe_resize_id=mom2

Try It

Use an online graphing tool to plot the equation $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$. Adjust the values you use for $a,b$ to values between $1, 20$. 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 $\left(h,k\right)$ other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms $\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ for horizontal hyperbolas, and $\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}-\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}=1$ 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 $\left(h, k\right)$ as it was in the ellipse.

How To: Given a general form for a hyperbola centered at $\left(h,k\right)$, 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 $\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$, then
• the transverse axis is parallel to the x-axis
• the center is $\left(h,k\right)$
• the coordinates of the vertices are $\left(h\pm a,k\right)$
• the coordinates of the co-vertices are $\left(h,k\pm b\right)$
• the coordinates of the foci are $\left(h\pm c,k\right)$
• the equations of the asymptotes are $y=\pm \frac{b}{a}\left(x-h\right)+k$
• If the equation is in the form $\dfrac{{\left(y-k\right)}^{2}}{{a}^{2}}-\dfrac{{\left(x-h\right)}^{2}}{{b}^{2}}=1$, then
• the transverse axis is parallel to the y-axis
• the center is $\left(h,k\right)$
• the coordinates of the vertices are $\left(h,k\pm a\right)$
• the coordinates of the co-vertices are $\left(h\pm b,k\right)$
• the coordinates of the foci are $\left(h,k\pm c\right)$
• the equations of the asymptotes are $y=\pm \frac{a}{b}\left(x-h\right)+k$
• Solve for the coordinates of the foci using the equation $c=\pm \sqrt{{a}^{2}+{b}^{2}}$.
• 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 $a\left(x^2+bx\right)$, add $\left(\dfrac{b}{2}\right)^2$ inside the parentheses, then subtract $a\left(\dfrac{b}{2}\right)^2$ to counteract the change you made.

If completing the square on one side of an equation, you may either subtract the value of $a\left(\dfrac{b}{2}\right)^2$ 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

$\qquad a\left(x^2+bx\right)$

$=a\left(x^2+bx+ \left(\dfrac{b}{2}\right)^2\right)-a\left(\dfrac{b}{2}\right)^2$

$=a\left(x+ \dfrac{b}{2}\right)^2-a\left(\dfrac{b}{2}\right)^2$

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

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

Try It

Graph the hyperbola given by the standard form of an equation $\dfrac{{\left(y+4\right)}^{2}}{100}-\dfrac{{\left(x - 3\right)}^{2}}{64}=1$. 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 $\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$.

Adjust the values you use for $a,b$ to values between $1, 20$, and the variables $h,k$ to numbers between $10,10$. 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