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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
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