How do you find the equation for a hyperbola centered at the origin with a horizontal transverse axis of lengths 8 units and a conjugate axis of lengths 6 units?

1 Answer
Dec 8, 2016

Answer:

The equation is: #(x - 0)^2/4^2 - (y - 0)^2/3^2 = 1#

Explanation:

Using this reference Conics: Hyperbola, the general equation for a hyperbola with a transverse horizontal axis is:

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

where #(h, k)# is the center, #2a# is the length of transverse axis, and #2b# is the length of the conjugate axis.

We are given that the center is, #(0, 0)#; substitute this into equation [1]:

#(x - 0)^2/a^2 - (y - 0)^2/b^2 = 1" [2]"#

We are given that the length of the horizontal transverse axis is 8; this allows us to write this equation:

#2a = 8#

#a = 4#

Substitute 4 for "a" in equation [2]:

#(x - 0)^2/4^2 - (y - 0)^2/b^2 = 1" [3]"#

We are given that the length of the horizontal transverse axis is 6; this allows us to write this equation:

#2b = 6#

#b = 3#

Substitute 3 for "b" in equation [3]:

#(x - 0)^2/4^2 - (y - 0)^2/3^2 = 1" [4]"#