Question 616031: Can someone help me with this question for my review?
x^2/36-y^2/16=1
center:
vertices:
conjugate points:
foci:
asymptotes:
I don't know how to start, it's overwhelming X_X IM STUCK....
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! x^2/36-y^2/16=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of the center
For given equation:x^2/36-y^2/16=1
center: (0,0)
..
a^2=36
a=√36=6
vertices: (0±a,0)=(0±6,0)=(-6,0) and (6,0)
..
b^2=16
b=√16=4
conjugate points: (0,0±b)=(0,0±4)=(0-4) and (0,4)
..
c^2=a^2+b^2=36+16=52
c=√52≈7.2
foci: (0±c,0)=(0±7.2,0)=(-7.2,0) and (7.2,0)
..
slope of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±4/6=±2/3
asymptotes are straight lines that intersect at the center.
Equation for asymptote with negative slope:
y=-2x/3+b
since y-intercept is at 0, b=0
equation of asymptote: y=-2x/3
..
Equation for asymptote with positive slope:
y=2x/3+b
since y-intercept is at 0, b=0
equation of asymptote: y=2x/3
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