Question 597522: I'm so lost on parabola and hyperbola and don't know where to start on these problems
1) (y-3)^2-4(x-4)^2=16
it is asking for the standard form of this equation, center, vertices, foci ,asymptotes
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! (y-3)^2-4(x-4)^2=16
it is asking for the standard form of this equation, center, vertices, foci ,asymptotes
**
(y-3)^2-4(x-4)^2=16
divide by16
(y-3)^2/16-(x-4)^2/4=1
This is an equation of a hyperbola with vertical transverse axis (y-term comes first) of the standard form: (y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of center
For given equation:(y-3)^2/16-(x-4)^2/4=1
center: (4,3)
a^2=16
a=√16=4
b^2=4
b=√4=2
c^2=a^2+b^2=16+4=20
c=√20≈4.5
..
vertices: (4,3±a)=(4,3±4)=(4,-1) and (4,7)
foci: (4,3±c)=(4,3±4.5)=(4,-1.5) and (4,7.5)
..
Asymptotes are straight lines of the standard form y=mx+b, m=slope, b=y-intercept.
Asymptotes also go thru the center.
slope of asymptotes=±a/b=±4/2=±2
Equations of asymptotes:
y=2x+b
solving for b using coordinates of center which are on the line
3=2*4+b
b=-5
equation: y=2x-5
..
y=-2x+b
solving for b using coordinates of center which are on the line
3=-2*4+b
b=11
equation: y=-2x+11
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