# 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 =

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> 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 =       Log On

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 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(6469)   (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