How do you graph #x^2/64-y^2=1# and identify the foci and asympototes?
1 Answer
May 27, 2017
For a hyperbola of the form:
The foci are at
The asymptotes are:
Explanation:
Matching the given equation,
with equation [1], we observe that
With these values the foci are at:
Simplify:
The asymptotes are:
Simplify:
The graph is:
graph{(x-0)^2/8^2-(y-0)^2/1^2=1 [-10, 10, -5, 5]}