Question
find a hyperbola equation with vertices (o,+-4) and passes through the point (3,-5)
Answers
Steve
we know that the center is at (0,0) and the axis is vertical, so
y^2/a^2 - x^2/b^2 = 1
If the distance between the vertices is 2a = 8, then 4^2 + b^2 = c^2
Since (3,-5) is on the graph,
25/a^2 - 9/b^2 = 1
a=4, so
25/16 - 9/b^2 = 1
b^2 = 16
y^2/16 - x^2/16 = 1
See the graph at
http://www.wolframalpha.com/input/?i=plot+y%5E2%2F16+-+x%5E2%2F16+%3D+1
y^2/a^2 - x^2/b^2 = 1
If the distance between the vertices is 2a = 8, then 4^2 + b^2 = c^2
Since (3,-5) is on the graph,
25/a^2 - 9/b^2 = 1
a=4, so
25/16 - 9/b^2 = 1
b^2 = 16
y^2/16 - x^2/16 = 1
See the graph at
http://www.wolframalpha.com/input/?i=plot+y%5E2%2F16+-+x%5E2%2F16+%3D+1