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Question 421165: find the center, vertices, foci and asymptotes of the hyperbola 1x^2-y^2-8x+2y-1=0 and draw a graph.
This is what I tried.
4x^2-8x+(-8/2)^2-y2+2y+(3/2)^2
4x^28x+16-4^2+2y+1=1+16+1
(2x-4)^2-(y+1)^2=18
This is driving me made. Thanks for the help.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find the center, vertices, foci and asymptotes of the hyperbola 1x^2-y^2-8x+2y-1=0 and draw a graph.
This is what I tried.
4x^2-8x+(-8/2)^2-y2+2y+(3/2)^2
4x^28x+16-4^2+2y+1=1+16+1
(2x-4)^2-(y+1)^2=18
..
In the given expression the first term is 1x^2; whereas, in the expression you worked with, the first term is 4x^2. I will work this problem assuming the latter is correct.
..
4x^2-y^2-8x+2y-1=0
Standard form of hyperbola: (x-h)2/a^2-(y-k)^2/b^2=1
completing the square(When doing this, factor out 4 so x^2 will have a coefficient of 1. I believe you forgot to do this when you tried.)
=4(x^2-2x+1)-(y^2-2y+1)=1+4-1=4
=4(x-1)^2-(y-1)^2=4
divide by (4)
=(x-1)^2/1-(y-1)^2/4=1
This is a hyperbola with center at (1,1) and horizontal transverse axis,that is, hyperbola opens sideways.
a^2=1
a=1
b^2=4
b=2
c^2=a^2+b^2
c=sqrt(1+4)=sqrt(5)=2.24..
vertices:(0,1) and (2,1)
foci:(3.24,1) and (-1.24,1)
Equations of asymptotes: (Use form, y-mx+b, with m=+-(b/a) of the hyperbola, and finding b of the equation, by using the coordinates of the center thru which the asymptotes pass.)
y=+-(b/a)(x-1)+1
y=+-(2/1)(x-1)+1
y=+-2(x-1)+1
See graph of the hyperbola below:
..
y=1+-(4(x-1)^2-4)^.5
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