Saturday
March 25, 2017

Post a New Question

Posted by on .

Find the equation of a hyperbola with foci of

(0,8), and (0,-8)

and Asymptotes of

y=4x and y=-4x


Please show work!!!

  • Adv. Algebra- Conics - ,

    In a hyperbola with centre at the origin the equation of the asymptote is y = ±(b/a)x.
    so b/a = 4 or b = 4a
    Also c = 8
    then in a^2 + b^2 = c^2 for a hyperbola
    a^2 + 16a^2 = 64
    17a^2 = 64
    a = 8/√17
    then b=32/√17

    using x^2/a^2 = y^2/b^2 = -1

    x^2/(64/17) - y^2/(32/17) = -1

    17x^2 /64 - 17y^2 /64 = -1

  • Adv. Algebra- Conics - ,

    correction:
    using x^2/a^2 = y^2/b^2 = -1 should say

    using x^2/a^2 - y^2/b^2 = -1

  • Adv. Algebra- Conics - ,

    Wow it's that easy. I was making it harder that it is.
    I think that's how you set it up, but wouldn't 32^2 be 1024

  • Adv. Algebra- Conics - ,

    <..but wouldn't 32^2 be 1024>

    of course, good for you for catching that.

    let's blame it on a "senior moment"

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question