SOLUTION: I am having trouble using the point slope formula to find the equations of the asymptotes of ((x+3)^2/16)-((y+2)^2/25)=1. I know the slope is +or-5/4. I'm just not sure where to g

Question 65049: I am having trouble using the point slope formula to find the equations of the asymptotes of ((x+3)^2/16)-((y+2)^2/25)=1. I know the slope is +or-5/4. I'm just not sure where to go from there. I also need some help finding the equation of the hyperbola described as having the foci at (5,0) and (0,-5) and the y-int. at (0,4) and (0,-4). Thank you very much.
Answer by cristiana(10) (Show Source): You can put this solution on YOUR website!
((x+3)^2/16)-((y+2)^2/25)=1 - equation of a hyperbola
(-3, -2) - the center
Slope = 5/4 or -5/4, just like you said.
Now, both asymptotes are going through the center, which means that we have the point we need to write the point-slope form of their equations.
Therefore, the point-slope of an equation going through (-3,-2) and having the slope 5/4 or -5/4 is: y+2 = 5/4(x+3), respectively y+2=-5/4(x+3). I'll let you do the math.
Cheers,
Cristiana
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