Asked by PLZ Help
What is the vertices of the hyperbola with the equation (16x^2) - (4y^2) = 64?
Answers
Answered by
oobleck
recall that x^2/a^2 - y^2/b^2 = 1
has vertices at (±a,0)
Your hyperbola can be written as
x^2/4 - y^2/16 = 1
so its vertices are at (±2,0)
has vertices at (±a,0)
Your hyperbola can be written as
x^2/4 - y^2/16 = 1
so its vertices are at (±2,0)
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