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what is the equation of a hyperbola with vertex (6,4) and (6,8) and foci(6,0 and 6,12.

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i really need an answer for this quickly so any help would be appreciated:-)
asked Mar 17, 2014 in ALGEBRA 2 by harvy0496 Apprentice

1 Answer

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The vertices of the hyperbola are (6, 4) and (6, 8) and its foci are (6, 0) and (6, 12).

The standard form of the equation of a hyperbola with center (h, k) (where a and b are not equals to 0) is (x - h)2/a2 - (y - k)2/b2 = 1 (Transverse axis is horizontal) or (y - k)2/a2 - (x - h)2/b2 = 1 (Transverse axis is vertical).

The vertices and foci are, respectively a and c units from the center (h, k) and the relation between a, b and c is b2 = c2 - a2.

Since the x - coordinate is constant in the vertices and foci, this is a vertical hyperbola.

The center of the hyperbola lies at the midpoint of its vertices or foci.

So, the center (h, k) = [ (x₁ + x₂)/2, (y₁ + y₂)/2 ] = [ (6 + 6)/2, (4 + 8)/2 ] = (6, 6).

To find the value of a, find the distance (positive) between the center (6, 6) and a vertex (6, 4).

d = sqrt [ (x₂ - x₁)2 + (y₂ - y₁)2 ]

d = sqrt [ (6 - 6)2 + (4 - 6)2 ]

d = sqrt [4] = 2

This distance is 2, so a = 2.

To find the value of c, find the distance between the center (6, 6) and a focus (6, 0).

d = sqrt [ (6 - 6)2 + (0 - 6)2 ]

d = sqrt [36] = 6.

This distance is 6, so c = 6.

Find the value of  b :

b2 = c2 - a2

b2 = 36 - 4 = 32.

b2 = 32

b = √32

The hyperbola equation is (y - 6)2/22 - (x - 6)2/(√32)2 = 1.

answered Mar 27, 2014 by steve Scholar
edited Mar 27, 2014 by steve

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