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Question 64531: Could someone help, Locate the vertices of the following hyperbola:
(x-5)^2/81 - (y-7)^2/49 =1
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Could someone help, Locate the vertices of the following hyperbola:
(x-5)^2/81 - (y-7)^2/49 =1
IT IS ALREADY IN STD.FORM
[(X-H)^2/A^2] - [(Y-K)^2/B^2]=1
DO YOU MEAN FOCI? OR CENTER? WE DO NOT SAY VERTICES OF HYPERBOLA.
SEE THE FOLLOWING PROBLEM AND DO.IF STILL IN DIFFICULTY PLEASE COME BACK
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refer to the hyperbola 36x 2 - 100y 2 - 72x + 400y = 3964.
36(X-1)^2-100(Y-2)^2=3964+36-400=3600
[(X-1)^2/10^2]-[(Y-2)^2/6^2]=1
THIS IS IN STD. FORM
[(X-H)^2]-[(Y-K)^2/B^2]=1.....H=1...K=2....A=10....B=6.....
CENTER IS (H,K)=(1,2)
ECCENTRICITY=E=SQRT[(A^2+B^2)/A^2]=SQRT(136/100)=1.1662
FOCI ARE [H+ AND - AE,K]=[12.662,2] AND [-10.662,2]
LENGTH OF TRANSVERSE AXIS = 2A=20
ASYMPTOTES ARE GIVEN BY EQN.
[(X-1)^2/10^2]-[(Y-2)^2/6^2]=0............OR..........
(X-1)/10=(Y-2)/6.... AND.....(X-1)/10=-(Y-2)/6
3X-3=5Y-10.............AND.......3X-3=-5Y+10
3X-5Y+7=0...........AND...........3X+5Y-13=0
The center of the hyperbola is
(2,1)
(1,2)...........CORRECT
(-1,2)
(1,-2)
The foci of the hyperbola are at the points
(10.662, 2) and (-12.662, 2)
(-10.662, -2) and (12.662, -2)
(-10.662, 2) and (12.662, 2)...................CORRECT
(2, -10.662) and (2, 12.662)
The length of the transverse axis is
5 units
10 units
20 units...................CORRECT
none of the above
The equations of the two asymptotes are
y = (3/5)x + 7/5 and...............CORRECT
y = (-3/5)x + 13/5.....................CORRECT
y = (-3/5)x + 7/5 and
y = (3/5)x + 13/5
y = (5/3)x + 7/5 and
y = (-5/3)x + 13/5
y = (3/5)x + 5/7 and
y = (-3/5)x + 5/13
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