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Question 1047763: sketch the hyperbola and label the vertices,foci and asymptotes of x²-4y²+2x+8y-7=0
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! x²-4y²+2x+8y-7=0
complete Squares
(x+1)^2 - 4(y-1)^2 = 7+ 1 - 4
Opens left and right along y = 1
C(-1,1) , a = 2
V(1,1) and (-3,1)
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foci
sqrt(4 + 1) = sqrt5
F(-1 + √5, 1) and F(-1-√5, 1)
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Asymptotes
m = ± 1/2
y - 1 = (1/2)(x+1)
y = (1/2)x + 3/2
0r
y - 1 = -(1/2)(x+1)
y= -(1/2)x + 1/2
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Need to Know
Standard Form of an Equation of an Hyperbola opening right and left is:
 with C(h,k) and vertices 'a' units right and left of center, 2a the length of the transverse axis. e = c/a.
Foci are  = c- units right and left of center along y = k
& Asymptotes Lines passing thru C(h,k), with slopes m = ± b/a
ow:
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