# SOLUTION: 2y^2-x^2=18. Solve to find asymptopes for a hyperbola. I know that the form for the hyperbola is y^2/a^2-x^2/b^2=1. I tried dividing everything by 18 to get 1 to the right of =, th

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: 2y^2-x^2=18. Solve to find asymptopes for a hyperbola. I know that the form for the hyperbola is y^2/a^2-x^2/b^2=1. I tried dividing everything by 18 to get 1 to the right of =, th      Log On

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 Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 35064This question is from textbook college algebra : 2y^2-x^2=18. Solve to find asymptopes for a hyperbola. I know that the form for the hyperbola is y^2/a^2-x^2/b^2=1. I tried dividing everything by 18 to get 1 to the right of =, that gives 2y^2/18-x^2/18 or y^2/9-x^2/18=1. I know 9 is 3^2 but 18 is not a perfect square, so I don't know what to do with it. This question is from textbook college algebra Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!YOU GOT IT ALMOST BUT STOPPED AT THE END JUST A FRACTION SHORT OF FINISH LINE.SEE MY COMMENTS BELOW. 2y^2-x^2=18. Solve to find asymptopes for a hyperbola. I know that the form for the hyperbola is y^2/a^2-x^2/b^2=1. VERY GOOD I tried dividing everything by 18 to get 1 to the right of =, that gives 2y^2/18-x^2/18 or y^2/9-x^2/18=1. EXCELLENT I know 9 is 3^2 PERFECT but 18 is not a perfect square, so I don't know what to do with it. NOTHING... LEAVE IT LIKE THAT... WRITE AS {SQRT(18)}^2 IF YOU WANT ..BUT THAT IS ALSO NOT NEEDED.. SO ASYMPTOTES COMBINED EQN.IS.... y^2/9-x^2/18=K SINCE THESE ARE PASSING THROUGH ORIGIN..WE GET 0-0=K...OR....K=0...SO PAIR OF ASYMPTOTES IS y^2/9-x^2/18=0...OR.... 18Y^2-9X^2=0...OR... 2Y^2-X^2=0....OR..... {YSQRT(2)+X}{YSQRT(2)-X}=0...OR.... YSQRT(2)=X AND YSQRT(2)=-X