Write in standard form of the hyperbola with the vertices of (1,5) and (1,11) and foci at (1,4) and (1,12)

Question

Standard form equation 

Answer 1

Notice that the vertices and foci have common x-values, x=1, which tells us that the graph of this hyperbola has a vertical transverse axis. The standard form of the equation of a hyperbola with a vertical transverse axis is as follows:

                          (y - k)²/a²  -  (x - h)²/b²  =  1 

     where (h, k) is the center of the hyperbola, the vertices are at (h, k+a) and (h, k-a), and the foci are at (h, k+c) and (h, k-c).

Vertices:     (1, 5) = (h, k+a)   ==>   k + a = 5   ==>   k = 5 - a

                  (1, 11) = (h, k-a)   ==>   k - a = 11   ==>   k = 11 + a

     5 - a = 11 + a   ==>   2a = -6   ==>   a = -3   ==>   a² = 9

Foci:     (1, 4) = (h, k+c)   ==>   k + c = 4   ==>   k = 4 - c

            (1, 12) = (h, k-c)   ==>   k - c = 12   ==>   k = 12 + c

     4 - c = 12 + c   ==>   2c = -8   ==>   c = -4   ==>   c² = 16

Find b using the following formula:     b² = c² - a²

          b² = 16 - 9   ==>   b² = 7   ==>   b = √7

Solve for k by plugging in appropriate variable into one of the equations determined for k: 

     k = 5 - a     ==>     k = 5 - (-3) = 5 + 3 = 8     ==>     k = 8

Thus, given that   h = 1 ,   k = 8 ,   a² = 9 ,   and   b² = 7 , the equation of the hyperbola is as follows:

               (y - 8)²/9  -  (x - 1)²/7  =  1

Answer 2

The hyperbola is North-South opening , so its equation has the form
(y−k)²/a²-(x−h)²/b² = 1
with
center (h, k), in this case (1,8), i.e. the mid point between vertices or foci.
vertices (h, k±a), i.e. (1, 8±3), a = distance between vertex and center = 3
foci (h, k±c), i.e. (1, 8±4), c = distance between focus and center = 4

Because c² = a² + b²,   b²=4²-3²=7

The equation becomes (y−8)²/9-(x−1)²/7 = 1