How do you write a standard form equation for the hyperbola with foci at (3,11) and (3,-5) and vertices at (3,7) and (3,-1)?

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Answer 1 · 2017-12-26T16:30:33

#(y-3)^2/16 -(x-3)^2/48 = 1#

The midpoint of the segment connecting the vertices (or the foci) is the center, #(h,k)\rightarrow(3,3)#.

The distance from the center to a focus is #c\rightarrow c=8#.
The distance from the center to a vertex is #a\rightarrow a=4#.

In a hyperbola we have the relationship#c^2=a^2+b^2# and we know both #a# and #c# so we can solve for #b^2 \rightarrowb^2=64-16 = 48#.

Since the vertices are on the same vertical line, #y# comes first in the equation, so the equation has the form:

#(y-k)^2/a^2 -(x-h)^2/b^2 = 1#

We can fill in to get:

#(y-3)^2/16 -(x-3)^2/48 = 1#