How do you write the equation for a hyperbola given vertices (-5,0) and (5,0), conjugate axis of lengths 12 units?

1 Answer
Nov 11, 2016

Answer:

Please see the explanation for steps leading to the equation:

#(x - 0)^2/5^2 - (y - 0)^2/6^2 = 1#

Explanation:

Hyperbola reference

The two vertices have the same y coordinate, therefore, the hyperbola is the horizontal transverse axis type. The standard equation for the horizontal transverse axis type is:

#(x - h)^2/a^2 - (y - k)^2/b^2 = 1#

We know that the general form for the vertices of this type are:

#(h + a, k)# and #(h -a, k)#

Using the given vertices #(-5, 0)# and #(5,0)#, we can write the following equations:

#k = 0#
#h + a = 5#
#h - a = -5#

Solve for h and a:

#a = 5#
#h = 0#

From the reference, the length of the conjugate axis is equal to 2b:

#2b = 12#

#b = 6#

We know the values of h, k, a, and b, substitute them into the general form:

#(x - 0)^2/5^2 - (y - 0)^2/6^2 = 1#