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Question 451835: what is the standard form of x^2-4x-4y^2+8y-100=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! what is the standard form of x^2-4x-4y^2+8y-100=0
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Completing the square is the first for finding out what kind of conic this equation is:
(x^2-4x+4)-4(y^2-2y+1)=100
(x-2)^2-4(y-1)^2=100+4-4=100
divide by 100
(x-2)^2/100-(y-1)^2/25=1
Standard form of a hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
Standard form of a hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
Note that the only difference between the two forms is the interchange of (y-k)^2 with (x-h)^2.
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Given hyperbola is of the first form with a horizontal transverse axis (hyperbola opens sideways) and center at (2,1).
a^2=100
a=10
length of transverse axis=2a=20
b^2=25
b=5
Length of conjugate axis=2b=10
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