find the vertices of the hyperbola

(y^2/4)-(x^2/16)=1
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Given (y^2/4)-(x^2/16)=1
Vertices of hyperbola are (0,-2) and (0,2)

Graph - http://www.wolframalpha.com/input/?i=%28y^2%2F4%29-%28x^2%2F16%29%3D1

Online Geometry Help - http://math.tutorcircle.com/geometry/

by Level 8 User (30.1k points)

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