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Write an equation for the hyperbola that satisfies the given set of conditions. Vertices (16,0) and (-16,0) conjugate axis of length 16 units.

User Kayes
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1 Answer

4 votes
vertices (23 , 0) and (-23 , 0) ==>
center (0 , 0), a = 23
if conjugate axis = 6 = 2b, then b = 3

for hyperbola that opens L and R (as shown by the vertices):
(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1 , where (h,k) is the center

x^2 / 23^2 - y^2 / 9 = 1
(center (0,0))

2nd one is similar
(center (0,0), a = 18 , b = 7)

3rd: standard form for hyperbola that opens up and down:
(y - k)^2 / a^2 + (x - h)^2 / b^2 = 1

a = 17, b = 5, (h,k) = (0,0)
center is always halfway between the vertices
y^2 / 17^2 - x^2 / 5^2 = 1

4th is similar to 3rd,
center (0,0) , a = 16 , b = 7

b = half of the conjugate axis

Hope it helps!!!
User Stefan Stefanov
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