SOLUTION: Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ± 1 divided by 4 x. (5 points) y squared over 16 minus x squared over 64 =

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Question 1138308: Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ± 1 divided by 4 x. (5 points)

y squared over 16 minus x squared over 64 = 1

y squared over 16 minus x squared over 256 = 1

y squared over 256 minus x squared over 16 = 1

y squared over 64 minus x squared over 4 = 1

Answer by MathLover1(20850) (Show Source):
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Find an equation in standard form for the hyperbola with vertices at (, ±)
and asymptotes at = ±

standard form:
since the vertices are at (0, ±4), that means the parabola opens up and down, and (the distance from center to vertex)
the center will be halfway between the vertices at:(, ) = (, )
the hyperbola will look like:

since given asymptotes at = ± , the slope of the vertices will be ± , and we have
± = ±=>
substituting in :