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Question 74845: i am working on hyperbola's and im not sure if i am understanding it yet. could see if i got the right answer to this one? thanks for any help. this is from a problem we had in class.
The center of the hyperbola 9x^2-16y^2-18x+32y-151=0
the answer i got was (1,1)
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! i am working on hyperbola's and im not sure if i am understanding it yet. could see if i got the right answer to this one? thanks for any help. this is from a problem we had in class.
The center of the hyperbola 9x^2-16y^2-18x+32y-151=0
[(3X)^2-2(3X)(3)+3^2-3^2]-[(4Y)^2-2(4Y)(4)+4^2-4^2]-151=0
(3X-3)^2-(4Y-4)^2-9+16-151=0
9(X-1)^2-16(Y-1)^2=144
[(X-1)^2/(12/3)^2]-[(Y-1)^2/(12/4)^2]=1
(X-1)^2/4^2 - (Y-1)^2/3^2 =1
THIS IS AS PER STD. EQN.
(X-H)^2/A^2 - (Y-K)^2/B^2=1
H=1
K=1
A=4
B=3
CENTER OF THE HYPERBOLA =(H,K)=(1,1)
the answer i got was (1,1)....CORRECT
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