How do you find the standard form of the equation of the hyperbola given the properties foci #(+-5,0)#, length of the conjugate axis 6?

1 Answer
Jul 1, 2017

Answer:

Please see the explanation.

Explanation:

Because the foci are given to be #(-5,0)# and #(5,0)#, we know that the hyperbola is the horizontal transverse axis type and has the general equation:

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

We know that #2b# is the length of the conjugate axis; this allows us to write the equation:

#2b = 6#

#b=3#

Substitute the value for "b" into equation [1]:

#(x-h)^2/a^2-(y-k)^2/3^2=1" [2]"#

We know that the general forms for the foci are:

#(h-sqrt(a^2+b^2),k)# and #(h+sqrt(a^2+b^2),k)#

This allows us to write the following equations:

#k=0" [3]"#
#-5 = h-sqrt(a^2+3^2)" [4]"#
#5 = h+sqrt(a^2+3^2)" [5]"#

Substitute the value for k into equation [2]:

#(x-h)^2/a^2-(y-0)^2/3^2=1" [6]"#

Find the value of h by adding equations [4] and [5]:

#2h = 0#

#h = 0#

Substitute into equation [6]:

#(x-0)^2/a^2-(y-0)^2/3^2=1" [7]"#

Use equation [5] to find the value of "a":

#5 = 0+sqrt(a^2+3^2)#

#25 = a^2+9

#a^2 = 16#

#a = 4#

Substitute the value of "a" into equation [7]

#(x-0)^2/4^2-(y-0)^2/3^2=1" [8]"#

Equation [8] is the equation of the hyperbola.