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
Douglas K.
Nov 12, 2016

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

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