SOLUTION: the question says find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola (x+1)sqared -(y-3)sqared=4
please help im so lost
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Question 613530: the question says find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola (x+1)sqared -(y-3)sqared=4
please help im so lost
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
the question says find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola (x+1)sqared -(y-3)sqared=4
**
(x+1)^2-(y-3)^2=4
(x+1)^2/4-(y-3)^2/4=1This 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+1)^2/4-(y-3)^2/4=1
center: (-1,3)
a^2=b^2=4
a=b=2
vertices: (-1±a,3)=(-1±2,3)=(-3,3) and (1,3)
..
foci:
c^2=a^2+b^2=4+4=8
c=√8≈2.8
foci: (-1±c,3)=(-1±2.8,3)=(-3.8,3) and (1.8,3)
..
Equations of Asymptotes:
Asymptotes are straight lines that go thru the center.
Slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±2/2=±1
Equation of asymptotes: y=±mx+b, m=slope, b=y-intercept
..
asymptote with negative slope:
y=-x+b
solve for b using (x,y) coordinates of center(-1,3)
3=1+b
b=2
equation: y=-x+2
..
asymptote with positive slope:
y=x+b
solve for b using (x,y) coordinates of center(-1,3)
3=-1+b
b=4
equation: y=x+4
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