Post a New Question

Algebra 2 Mannn

posted by on .

Can someone please explain to me how to get the 16 in this equation. I get how to get everything else but i don't know how to get the 16.

A hyperbola with foci: (0,5) (0,-5)and vertices: (0,3) (0,-3). So i got y^2/9-x^2/16=1. but how do you get the 16?

  • Algebra 2 Mannn - ,

    center is at (0,0) so the form is:
    y^2/a^2 -x^2/b^2 = 1

    distance center to focus = 5 = sqrt(a^2+b^2)
    distance center to vertex = 3 = a
    so
    5^2 = (3^2 + b^2)
    25 = 9 + b^2
    b^2 = 16

  • Algebra 2 Mannn - ,

    thanks man

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question