# SOLUTION: Find the standard form of the equation of the hyperbola with the given characteristics. asymptotes:y=+-4x a.(y^2)/(16/17)-(x^2)/(256/17)=1 b.(x^2)/(16/17)-(y^2)/(256/17)=1

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: Find the standard form of the equation of the hyperbola with the given characteristics. asymptotes:y=+-4x a.(y^2)/(16/17)-(x^2)/(256/17)=1 b.(x^2)/(16/17)-(y^2)/(256/17)=1       Log On

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 Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 470818: Find the standard form of the equation of the hyperbola with the given characteristics. asymptotes:y=+-4x a.(y^2)/(16/17)-(x^2)/(256/17)=1 b.(x^2)/(16/17)-(y^2)/(256/17)=1 c.(y^2)/(16)-(x^2)/(16)=1 d.(x^2)/(16)-(y^2)/(16)=1 e.(x^2)/(256/17)-(y^2)/(16/17)=1Answer by lwsshak3(6469)   (Show Source): You can put this solution on YOUR website!Find the standard form of the equation of the hyperbola with the given characteristics. asymptotes:y=+-4x a.(y^2)/(16/17)-(x^2)/(256/17)=1 b.(x^2)/(16/17)-(y^2)/(256/17)=1 c.(y^2)/(16)-(x^2)/(16)=1 d.(x^2)/(16)-(y^2)/(16)=1 e.(x^2)/(256/17)-(y^2)/(16/17)=1 ** Standard form for hyperbola: (x-h)^2/a^2-(y-k)^2/b^2=1 b.(x^2)/(16/17)-(y^2)/(256/17)=1 This is a hyperbola with horizontal transverse axis. (opens sideways) Center: (0,0) a^2=16/17 a=4/√17 b^2=256/17 b=16/√17 Slope=b/a=(16/√17)/(4√17)=4 (matches given slope=±4) asymptotes: y=±4x ans:Equation b. is the correct choice see graph below as a visual check on the answer. .. y=±((17x^2/16-1)(256/17))^.5