Question 631670: where is the asymptotes, center, and equations of asymptotes. for this equation:
4y^2 - 16y - x^2 + 2x = 10
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! where is the asymptotes, center, and equations of asymptotes. for this equation:
4y^2 - 16y - x^2 + 2x = 10
complete the square
4(y^2-4y+4)-(x^2-2x+1) = 10+16-1
4(y-2)-(x-1)=25
This is an equation of a hyperbola with vertical transverse axis
Its standard form: 
For given equation:
center: (1,2)
a^2=25/4
a=√(25/4)=5/2
b^2=25
b=√25=5
asymptotes are straight lines that go thru the center: equation: y=mx+b, m=slope, b=y-intercept
slopes, m, of asymptotes for hyperbolas with vertical transverse axis=±a/b=±(5/2)5=±5/10=±1/2
..
asymptote with negative slope: y=-x/2+b
solve for b using coordinates of center (1,2)
2=-1/2+b
b=5/2
equation: y=-x/2+5/2
..
asymptote with positive slope: y=x/2+b
solve for b using coordinates of center (1,2)
2=1/2+b
b= 3/2
equation: y=-x/2+3/2
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