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

Standard form equation

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

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