since the asymptotes are y=4x, the center must be at (0,0)
The foci are on the y-axis, so the asymptotes are y = ±a/b x
since c=8 (from the foci)
and a/b = 4,
b^2 + (4b)^2 = 8^2
b = 8/√17
a = 32/√17
y^2/a^2 - x^2/b^2 = 1
y^2/(1024/17) - x^2/(64/17) = 1
or
17y^2/1024 - 17x^2/64 = 1
see the graph at
http://www.wolframalpha.com/input/?i=hyperbola+y%5E2%2F(1024%2F17)+-+x%5E2%2F(64%2F17)+%3D+1
find the standard form of the hyperbola, the center and vertices given the foci (0, -+8) and asymptotes at y=4x
1 answer
1 answer