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Question 610033: Graph each. Be sure to put in all foci and asymptotes 2 problems below thanks
(x+2)^2 divided by 9 - (y-1)^2 divided by 25 = 1
(x-2)^2=4(2)(y-1)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Graph each. Be sure to put in all foci and asymptotes 2 problems below thanks
(x+2)^2 divided by 9 - (y-1)^2 divided by 25 = 1
(x-2)^2=4(2)(y-1)
**
The easier problem first:
(x-2)^2=4(2)(y-1)
This is an equation of a parabola that opens upwards.
Its standard form: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex.
For given equation: (x-2)^2=4(2)(y-1)
vertex: (2,1)
axis of symmetry: x=2
p=2
4p=8
Focus: (2,3) (2 units above vertex on the axis of symmetry)
Asymptotes do not apply to parabolas.
See graph of given parabola below:
y=(x-2)^2/8+1
..
Second problem:
(x+2)^2/9-(y-1)^2/25=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 center
For given equation:(x+2)^2/9-(y-1)^2/25=1
center: (-2,1)
a^2=9
a=3
vertices: (-2±a,1)=(-2±3,1)=(-5,1) and (1,1)
b^2=25
b=5
c^2=a^2+b^2=9+25=34
c=√34≈5.8
Foci: (-2±c,1)=(-2±5.8,1)=(-7.8,1) and (3.8,1)
Asymptotes:
slope of asymptotes=±b/a=±5/3
Asymptotes are straight lines that go thru center of hyperbola.
Standard form of equation for straight line: y=mx+b, m=slope, b=y-intercept
Equation of asymptote with slope>0
y=5x/3+b
solve for b using (x,y) coordinates of center
1=5*-2/3+b
3=-10+3b
3b=13
b=13/3
equation:y=5x/3+13/3
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Asymptote with slope<0
y=-5x/3+b
solve for b using (x,y) coordinates of center
1=-5*-2/3+b
3=10+3b
3b=-7
b=-7/3
equation:y=-5x/3-7/3
see graph of given hyperbola below:
y=((25(x+2)^2/9)-25)^.5+1
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