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Question 65743: A hyperbola is the only one of the conic sections whose graph contains two separate curves.
True
False
Each of the remaining questions refer to the hyperbola 36x 2 - 100y 2 - 72x + 400y = 3964.
The center of the hyperbola is
(2,1)
(1,2)
(-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)
(2, -10.662) and (2, 12.662)
The length of the transverse axis is
5 units
10 units
20 units
none of the above
The equations of the two asymptotes are
y = (3/5)x + 7/5 and
y = (-3/5)x + 13/5
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
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! A hyperbola is the only one of the conic sections whose graph contains two separate curves.
True
False
TRUE SEE THE GRAPH BELOW
Each of the remaining questions 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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