# SOLUTION: Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y^/25-x^/64=1

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 Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 481495: Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y^/25-x^/64=1Answer by lwsshak3(6509)   (Show Source): You can put this solution on YOUR website!Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y^/25-x^/64=1 ** This is an equation of a hyperbola with vertical transverse axis of the standard form: (y-k)^2/a^2-(x-h)^2/b^2=1 For given equation: Center: (0,0) a^2=25 a=√25=5 Length of vertical transverse axis=2a=10 Vertices (end points of transverse axis): (0,0±a)=(0,0±5)=(0,5) and (0-5) .. b^2=64 b=√64=8 .. c^2=a^2+b^2=25+64=89 c=√89=9.4 .. Foci: (0,0±c)=(0,±√89)=(0,√89) and (0,-√89) .. Asymptotes: slope: ±a/b=±5/8 Equation of asymptotes which are straight lines of the standard form: y=mx+b, m=slope, b=y-intercept. Asymptotes also go thru center at (0,0) in this case, so y-intercept=0. Equation of asymptotes: y=±5x/8 See graph below as a visual check on answers: .. y=±(25+25x^2/64)^.5