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Question 982303: Find an equation in standard form for the hyperbola with vertices at (0, +/-10) and asymptotes at
y = +/- 5/4 x
My work, can someone please double check it?
Given: a = 10
The relationship between a, b, and c is
10² + b² = c²
The slope of the asymptotes is ±5/4
a = is a multiple of 5
b = is a multiple of 4
For this problem a = 10
hence, b = 8
10² + 8² = c²
164 = c²
2√41 = c
Summary
center = (0, 0)
a² = 100
b² = 64
c² = 164
transverse axis x = 0
equation of the hyperbola
[y²/10²] - [x²/8²] = 1
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
No. a is 10 because the vertices are (0,-10) and (0,10), not because it is a multiple of 5.
But the slopes of the asymptotes are  for any hyperbola and  . a is a multiple of 4, not 5. b is a multiple of 5.
Like I said before, solve for b
Cross multiply:
You can take it from here. Hint, just because every other example you have ever seen have a and b as integers, doesn't mean every hyperbola is that way. a and b can be any real numbers, they don't even have to be rational.
John
My calculator said it, I believe it, that settles it
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