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

Question

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

Answer 1

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