How do you find the standard form of the hyperbola that satisfies the given conditions: Vertices at #(0, +-2)#, foci at #(0, +-4)#?

1 Answer
Nov 12, 2016

Answer:

Please see the explanation for steps leading to the equation:

#(y - 0)^2/2^2 - (x - 0)^2/(sqrt(12))^2 = 1#

Explanation:

The y coordinate of the vertices is the one that changes, therefore, the hyperbola is the vertical transverse axis type. The general equation for this type is:

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

where:

The centerpoint is

#(h, k)#

The vertices are:

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

And the foci are:

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

Given that the vertices are #(0, -2)# and #(0, 2), we can write the following equations:

#h = 0" [1]"#
#k - a = -2" [2]"#
#k + a = 2" [3]"#

Use equations [2] and [3] to solve for a and k:

#2k = 0" [2 + 3]"#

#k = 0#

#2a = 4" [3 - 2]"#

#a = 2#

Given that one of foci is #(0, -4)#, allows us to write the following equation:

#-4 = k - sqrt(a^2 + b^2)#

Substitute 0 for k and 2 for a:

#-4 = 0 - sqrt(2^2 + b^2)#

Solve for b:

#16 = 4 + b^2#

#b^2 = 12#

#b = sqrt(12)#

Substitute all of the computed values into the general form:

#(y - 0)^2/2^2 - (x - 0)^2/(sqrt(12))^2 = 1#