Sunday, October 14, 2012

One of the 4 conic sections taught in algebra is the hyperbola.  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. 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, which will be one of two forms

x^2 - y^2  = 1                   y^2x^2 =1
a^2    b^2               or       a^2     b^2

The first thing we will do is find the intercepts of the hyperbola.. For the hyperbolas with the equation in the first form 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   x^2  - y^2 = 1
                                                                            25       9

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. Notice the graph of the hyperbola below.

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