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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.

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1 Answer

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

Comments

Hi Tamara. Equation of the Hyperbola with Center (h,k) and Focal Axis Parallel to y-axis has form: 
(y-k)2/a2 - (x-h)2/b2 = 1

Comment