Posted by **Tristan** on Saturday, May 11, 2013 at 10:47pm.

Find the equation of the asymptotes for the hyperbola defined by the equation:

((y + 4)^2 / 36) - ((x - 3)^2 / 25) = 1

- Algebra 2 -
**Reiny**, Sunday, May 12, 2013 at 12:41am
centre is (-4,3)

and the asymptotes go through the centre

slopes are -5/6

y = (5/6)x + b

at (-4,3)

3 = (5/6)(-4) + b

b = 39/5

y = (5/6)x + 39/5

or

y = (-5/6)x + b

3 = (-5/6)(-4) + b

b = -1/3

y = (-5/6)x - 1/3

## Answer this Question

## Related Questions

- Algebra - HELP! I am not sure how I would even start this problem or solve it. A...
- algebra - what is the equation of a hyperbola with one vertex at (6,5) and the ...
- College algebra - Find the standard form of the equation of the hyperbola with ...
- Adv. Algebra- Conics - Find the equation of a hyperbola with foci of (0,8), and...
- College Algebra. - Find the equation in standard form of the hyperbola that ...
- math - Find equations for the asymptotes of the hyperbola x2/9 - y2/81 = 1 ...
- math30 - The equation of the hyperbola is (x-3)^2/4 - (y+1)^2/16 =-1. What is ...
- Math/ Algebra - Find an equation for the hyperbola described: 1) Vertices at (0...
- precalculus - find an hyperbola equation with foci (0, +/-8) and asymptotes; y...
- pre-cal - find an equation for the hyperbola vertices (3,0),(3,4) asymptotes:y=2...