SOLUTION: The question I have for homework is Find an equation of the hyperbola having foci at (-1,1-sqrt52), and -1,1+sqrt52) and asymptotes at y=2/3x+5/3 and y=-2/3x+1/3 I know c =

Question 470703: The question I have for homework is
Find an equation of the hyperbola having foci at (-1,1-sqrt52), and -1,1+sqrt52) and asymptotes at y=2/3x+5/3 and y=-2/3x+1/3
I know c = sqrt of 52 and c^2 = 52 I am just unsure on how to find A and B. I know how to graph the asymptotes and I know the center is (-1,1)
I know I have to use 2/3 for A and B I just dont understand how. Any help would be greatly appreciated thanks..
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find an equation of the hyperbola having foci at (-1,1-sqrt52), and -1,1+sqrt52) and asymptotes at y=2/3x+5/3 and y=-2/3x+1/3
I know c = sqrt of 52 and c^2 = 52 I am just unsure on how to find A and B. I know how to graph the asymptotes and I know the center is (-1,1)
I know I have to use 2/3 for A and B I just dont understand how.
**
From given Foci coordinates you can see that this is a hyperbola with vertical transverse axis.
(y changes but x does not)
Standard form for hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being (x,y) coordinates of center
Slope, m, for asymptotes of this form=a/b
From given asymptote equations, m=�2/3
a/b=2/3
b=3a/2
b^2=9a^2/4
..
From Foci
c=√52
c^2=a^2+b^2
52=a^2+9a^2/4
208=4a^2+9a^2=/13a^2
a^2=208/13=16
b^2=9a^2/4=9*16/4=36
..
Equation:
(y-1)^2/16-(x+1)^2/36=1
Asymptotes:
2x/3+5/3
-2x/3+1/3
see graph below as a visual check on the answer
..
y=�(16+16(x+1)^2/36)^.5+1

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