Hyperbolas

Graph a Hyperbola with Center at (0, 0)

The last conic section we will look at is called a hyperbola. We will see that the equation of a hyperbola looks the same as the equation of an ellipse, except it is a difference rather than a sum. While the equations of an ellipse and a hyperbola are very similar, their graphs are very different.

We define a hyperbola as all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a focus of the hyperbola.

The line through the foci, is called the transverse axis. The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola. The midpoint of the segment joining the foci is called the center of the hyperbola. The line perpendicular to the transverse axis that passes through the center is called the conjugate axis. Each piece of the graph is called a branch of the hyperbola.

Again our goal is to connect the geometry of a conic with algebra. Placing the hyperbola on a rectangular coordinate system gives us that opportunity. In the figure, we placed the hyperbola so the foci $\left(\left(\text{−}c,0\right),\left(c,0\right)\right)$ are on the x-axis and the center is the origin.

The definition states the difference of the distance from the foci to a point $\left(x,y\right)$ is constant. So $\left|d_1-d_2\right|$ is a constant that we will call $2a$ so $\left|d_1-d_2\right|=2a.$ We will use the distance formula to lead us to an algebraic formula for an ellipse.

$\begin{array}{cccc}& & & \hfill |d_1-d_2|=2a\\ \text{Use the distance formula to find}{d}_1,d_2\hfill & & & \hfill \left|\sqrt{{\left(x-\left(-c\right)\right)}^2+{\left(y-0\right)}^2}-\sqrt{{\left(x-c\right)}^2+{\left(y-0\right)}^2}\right|=2a\\ \text{Eliminate the radicals.}\hfill & & & \\ \begin{array}{c}\text{To simplify the equation of the ellipse, we}\hfill \\ \text{let}{c}^2-a^2=b^2.\hfill \end{array}\hfill & & & \hfill \frac{x^2}{a^2}+\frac{y^2}{c^2-a^2}=1\\ \begin{array}{c}\text{So, the equation of a hyperbola centered at}\hfill \\ \text{the origin in standard form is:}\hfill \end{array}\hfill & & & \hfill \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\end{array}$

To graph the hyperbola, it will be helpful to know about the intercepts. We will find the x-intercepts and y-intercepts using the formula.

$\begin{array}{c}\begin{array}{cccccccccccccccc}& & & & & \mathit{\text{x}}\mathbf{\text{-intercepts}}\hfill & & & & & & & & & & \mathit{\text{y}}\mathbf{\text{-intercepts}}\hfill \end{array}\hfill \\ \begin{array}{cccccccccccccccccccc}& & & & & \hfill \frac{x^2}{a^2}-\frac{y^2}{b^2}& =\hfill & 1\hfill & & & & & & & & & & \hfill \frac{x^2}{a^2}-\frac{y^2}{b^2}& =\hfill & 1\hfill \\ \text{Let}y=0.\hfill & & & & & \hfill \frac{x^2}{a^2}-\frac{0^2}{b^2}& =\hfill & 1\hfill & & & & & \text{Let}x=0.\hfill & & & & & \hfill \frac{0^2}{a^2}-\frac{y^2}{b^2}& =\hfill & 1\hfill \\ & & & & & \hfill \frac{x^2}{a^2}& =\hfill & 1\hfill & & & & & & & & & & \hfill -\frac{y^2}{b^2}& =\hfill & 1\hfill \\ & & & & & \hfill x^2& =\hfill & a^2\hfill & & & & & & & & & & \hfill y^2& =\hfill & \text{−}{b}^2\hfill \\ & & & & & \hfill x& =\hfill & \text{±}a\hfill & & & & & & & & & & \hfill y& =\hfill & \text{±}\sqrt{\text{−}{b}^2}\hfill \end{array}\hfill \\ \begin{array}{cccccc}\text{The}x\text{-intercepts are}\left(a,0\right)\text{and}\left(\text{−}a,0\right).\hfill & & & & & \text{There are no}y\text{-intercepts.}\hfill \end{array}\hfill \end{array}$

The a, b values in the equation also help us find the asymptotes of the hyperbola. The asymptotes are intersecting straight lines that the branches of the graph approach but never intersect as the x, y values get larger and larger.

To find the asymptotes, we sketch a rectangle whose sides intersect the x-axis at the vertices $\left(\text{−}a,0\right),$ $\left(a,0\right)$ and intersect the y-axis at $\left(0,\text{−}b\right),$ $\left(0,b\right).$ The lines containing the diagonals of this rectangle are the asymptotes of the hyperbola. The rectangle and asymptotes are not part of the hyperbola, but they help us graph the hyperbola.

The asymptotes pass through the origin and we can evaluate their slope using the rectangle we sketched. They have equations $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x.$

There are two equations for hyperbolas, depending whether the transverse axis is vertical or horizontal. We can tell whether the transverse axis is horizontal by looking at the equation. When the equation is in standard form, if the x²-term is positive, the transverse axis is horizontal. When the equation is in standard form, if the y²-term is positive, the transverse axis is vertical.

The second equations could be derived similarly to what we have done. We will summarize the results here.

We will use these properties to graph hyperbolas.

How to Graph a Hyperbola with Center $\left(0,0\right)$

Graph $\frac{x^2}{25}-\frac{y^2}{4}=1.$

Solution

We summarize the steps for reference.

Sometimes the equation for a hyperbola needs to be first placed in standard form before we graph it.

Graph $4y^2-16x^2=64.$

Solution

	$4y^2-16x^2=64$
To write the equation in standard form, divide each term by 64 to make the equation equal to 1.	$\frac{4y^2}{64}-\frac{16x^2}{64}=\frac{64}{64}$
Simplify.	$\frac{y^2}{16}-\frac{x^2}{4}=1$
Since the y2-term is positive, the transverse axis is vertical. Since $a^2=16$ then $a=\text{±}4.$
The vertices are on the y-axis, $\left(0,\text{−}a\right),$ $\left(0,a\right).$ Since $b^2=4$ then $b=\text{±}2.$	$\left(0,\mathrm{-4}\right),$ $\left(0,4\right)$
Sketch the rectangle intersecting the x-axis at $\left(\mathrm{-2},0\right),$ $\left(2,0\right)$ and the y-axis at the vertices. Sketch the asymptotes through the diagonals of the rectangle. Draw the two branches of the hyperbola.

Graph a Hyperbola with Center at $\left(h,k\right)$

Hyperbolas are not always centered at the origin. When a hyperbola is centered at $\left(h,k\right)$ the equations changes a bit as reflected in the table.

How to Graph a Hyperbola with Center $\left(h,k\right)$

Graph $\frac{{\left(x-1\right)}^2}{9}-\frac{{\left(y-2\right)}^2}{16}=1$

Solution

We summarize the steps for easy reference.

Be careful as you identify the center. The standard equation has $x-h$ and $y-k$ with the center as $\left(h,k\right).$

Graph $\frac{{\left(y+2\right)}^2}{9}-\frac{{\left(x+1\right)}^2}{4}=1.$

Solution

Since the $y^2\text{-}$term is positive, the hyperbola opens up and down.
Find the center, $\left(h,k\right).$	Center: $\left(\mathrm{-1},\mathrm{-2}\right)$
Find a, b.	$a=3$ $b=2$
Sketch the rectangle that goes through the points 3 units above and below the center and 2 units to the left/right of the center. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices. Graph the branches.

Again, sometimes we have to put the equation in standard form as our first step.

Write the equation in standard form and graph $4x^2-9y^2-24x-36y-36=0.$

Solution

To get to standard form, complete the squares.
Divide each term by 36 to get the constant to be 1.
Since the $x^2\text{-}$term is positive, the hyperbola opens left and right.
Find the center, $\left(h,k\right).$	Center: $\left(3,\mathrm{-2}\right)$
Find a, b.	$\begin{array}{c}a=3\hfill \\ b=4\hfill \end{array}$
Sketch the rectangle that goes through the points 3 units to the left/right of the center and 2 units above and below the center. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices. Graph the branches.

Identify Conic Sections by their Equations

Now that we have completed our study of the conic sections, we will take a look at the different equations and recognize some ways to identify a conic by its equation. When we are given an equation to graph, it is helpful to identify the conic so we know what next steps to take.

To identify a conic from its equation, it is easier if we put the variable terms on one side of the equation and the constants on the other.