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April 18, 2015

Posted by **Paragon** on Monday, November 19, 2007 at 12:26am.

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

and Asymptotes of

y=4x and y=-4x

Please show work!!!

- Adv. Algebra- Conics -
**Reiny**, Monday, November 19, 2007 at 7:34amIn 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 -
**Reiny**, Monday, November 19, 2007 at 7:36amcorrection:

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 -
- Adv. Algebra- Conics -
**Paragon**, Monday, November 19, 2007 at 7:45amWow it's that easy. I was making it harder that it is.

I think thats how you set it up, but wouldn't 32^2 be 1024

- Adv. Algebra- Conics -
**Reiny**, Monday, November 19, 2007 at 8:14am<..but wouldn't 32^2 be 1024>

of course, good for you for catching that.

let's blame it on a "senior moment"

- Adv. Algebra- Conics -

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