In a high school course in advanced algebra or pre calculus, students will learn the four conic sections. One of the most confusing to understand is the hyperbola. At the conclusion of this article, you will understand how to identify the hyperbola and graph by its equation.

The shape of the hyperbola is similar to two parabolas that are equal distant apart opening in opposite directions along the same axis. If you flip each side of the parabola you will notice it will resemble an ellipse. In fact the definition of the ellipse and hyperbola are similar. A hyperbola is the set of all points in a single plane for which the difference of the distances from two fixed points is a constant. The only difference between the definitions and equations for the hyperbola and the ellipse is that the ellipse is sum instead of difference. If you look at half of a hyperbola, it looks similar to a parabola with the focus inside. The center of the hyperbolas is always the halfway point between the foci.

To graph a hyperbola with the center at the origin, we first need to know the equation of a hyperbola in standard form, x2/a2 – y2/b2 = 1 or y2/a2 – x2/b2 = 1. The first thing we will do is find the intercepts of the hyperbola. For the hyperbolas with the first equation above, the x – intercepts are at (-a, 0) and (a, 0) and the y – intercepts are at (0, -b) and (0, b). We can draw a rectangular box through these four coordinates and draw diagonal lines extended through the corners of the rectangle. These lines are the asymptotes of the hyperbola and are represented by the equations y = +/- (b/a)x. The foci can be found by using the equation a2 + b2 = c2. Once you find c, we move c units from the center (0, 0) in both directions along the axis where a is located, which is the x axis. We can now graph the parabola by drawing the curves through both vertices on the x- axis getting close to the asymptotes but never touching them.

**Example:** Graph the hyperbola with the equation x2/25 – y2/9 = 1.

We will first find the x and y intercepts. Recall that the x- intercepts are at (-a, 0) and (a, 0). Since a2 = 25, a = 5 and the x – intercepts are (-5, 0) and (5, 0). Next we will find the y- intercepts, which are at (0, -b) and (0, b). Since b2 = 9, b = 3 and the intercepts are at (0, -3) and (0, 3). At this point we plot these 4 coordinates and draw the rectangular box through these points. Next, draw the asymptotes, which are the 2 diagonals through the vertices of the rectangular box. The equations of the asymptotes are y = +/- (b/a)x, which is y = +/- (3/5)x. We can now find the foci by solving for c in the equation a2 + b2 = c2. Doing so, we get 25 + 9 = c2, 34 = c2 and c ≈ 5.83. Recall for the hyperbola in this form centered at the origin, the foci are c units in both directions from the center along the x axis. Therefore the foci are (-5.83, 0) and (5.83, 0). Place the foci on the graph and finish the graph by drawing the hyperbola through the vertices on the x- axis approaching the asymptotes but never touching them.

We will next consider the graph of the hyperbola with the equation where the y2 term is positive and the x2 term is negative. The process for graphing this is identical to in the previous example with the exception that the foci are along the y – axis and the vertices are (0, a), (0, -a), (-b, 0) and (b, 0) and the asymptotes are y = +/- (a/b)x.

**Example:** Graph the hyperbola with the equation y2/16 – x2/36 = 1.

We will first find the x and y intercepts. Recall that the x- intercepts for a hyperbola in this form are at (-b, 0) and (b, 0). Since b2 = 36, b = 6 and the x- intercepts are at (-6, 0) and (6, 0). Next we will find the y- intercepts which are at (0, -a) and (0, a). Since a2 = 16, a = 4 and the y- intercepts are at (0, -4) and (0, 4). At this point we plot the 4 coordinates and draw the rectangular box through these points. Next, draw the asymptotes, which are the 2 diagonals through the rectangular box. The equations of the asymptotes are y = +/- (a/b)x, which is y = +/- (4/6)x, simplified to y = +/- (2/3)x. We can now find the foci by solving for c in the equation a2 + b2 = c2. Doing so, we get 16 + 36 = c2, 52 = c2 and c ≈ 7.2. Recall for the hyperbola in this form centered at the origin, the foci are c units in both directions from the center along the y axis. Therefore the foci are (0, -7.2) and (0, 7.2). Place the foci on the graph and finish the graph by drawing the hyperbola through the vertices on the y- axis approaching the asymptotes but never touching them.

Note that it’s easy to remember which way to draw the hyperbola. If the x2 term is positive it opens horizontally (left and right). If the y2 term is positive, it opens vertically (up and down). It opens along the axis of the term that is positive.

Oftentimes, a hyperbola is not centered at the origin. In such cases, the point (h, k) is said to be the center. The equation changes to (x – h)2/a2 – (y – k)2/b2 = 1 or (y – k)2/a2 – (x – h)2/b2 = 1. The foci for a hyperbola of the first form are located at (h + c, k) and (h – c, k). Recall that the foci for the equation in this form are c units away from the center in both directions along the x- axis. The equations of the asymptotes are y = k +/- (b/a)(x – h).

the foci for a hyperbola of the second form are located at (h, k – c) and (h, k + c). Recall that the foci

for the equation in this form are c units away from the center in both directions along the y- axis. The equations of the asymptotes are y = k +/- (a/b)(x – h).