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# write the equation of a hyperbola in standard form whose center is (-2, -4), a focus at (-2, 6), and eccentricity of 5/4.

please show work so i can understand how you solved it. thanks!

### 1 Answer by Expert Tutors

Tamara J. | Math Tutoring - Algebra and Calculus (all levels)Math Tutoring - Algebra and Calculus (al...
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Given:     center = (h, k) = (-2, -4)

focus = (-2, 6)

eccentricity = c/a = (√(a2 + b2))/a = 5/4

First notice that both the center and the focus have the same x coordinate (x= -2), which means that the transverse axis is parallel to the y-axis (i.e., the transverse axis is vertical). From this we know that the standard form of this hyperbola is as follows:

(y - k)2/b2  -  (x - h)2/a2  =  1

Second, since the eccentricity is equal to c/a, then a=4 and c = 5. From this we can find b using the fact that c equal the square root of the sum of a2 and b2:

c = √(a2 + b2

5 = √(42 + b2)

5 = √(16 + b2)

Square both sides of this equation then subtract 16 from both sides:

25 = 16 + b2

25 - 16 = b2

9 = b2

3 = b

Using the values that we've found for a and b, as well as the center ((h, k) = (-2, -4)) given and the standard form of a hyperbola with a vertical transverse axis, we can generate the equation of the hyperbola in standard form:

(y - k)2/b2  -  (x - h)2/a2  =  1

(y - (-4))2/32  -  (x - (-2))2/42  =  1

(y + 4)2/9  -  (x + 2)2/16  =  1