# Hyperbola

• Mar 30th 2007, 04:12 PM
amanda1
Hyperbola
What would I need to do to find equations for the asymptotes of the hyperbola:

x[squared]/9 - y[squared]/81 = 1
• Mar 31st 2007, 07:08 AM
ecMathGeek
Quote:

Originally Posted by amanda1
What would I need to do to find equations for the asymptotes of the hyperbola:

x[squared]/9 - y[squared]/81 = 1

For future reference, we generally write x[squared] as x^2.

x^2/9 - y^2/81 = 1

a = sqrt(9) = 3
b = sqrt(81) = 9

I'm sure you already know how to draw this hyperbola, but if not, I'll explain the basic process. You first create a point at the center (h,k), which in this case is (0,0). From this, you would place points "a" units left and right of (h,k) and "b" units above and below (h,k). Around these points you would draw a box, and through the verticies of this box you would draw your asymptotes.

The slope of your asymptotes is (+/-)m = (+/-)(rise)/(run) = (+/-)b/a. Since both asymptotes go through the point (h,k), you can draw the equation of both lines as such:

y - k = (+/-)b/a*(x - h) --> y - 0 = (+/-)9/3*(x - 0) --> y = 3x AND y = -3x

NOTE: This is a horizontal hyperbola, which is why the slope is m = (+/-)b/a. If this were a vertical hyperbola, the slope would be m = (+/-)a/b.