SOLUTION: 1Find the vertex and focus of the parabola whose equation is 4y = x2 + 4. aV(0, 4), F(0, 3) bV(0, 1), F(0, 2) cV(4, 0), F(3, 0) dV(1, 0), F(2, 0) 2Find the center and

Question 486499: 1Find the vertex and focus of the parabola whose equation is 4y = x2 + 4.
aV(0, 4), F(0, 3)
bV(0, 1), F(0, 2)
cV(4, 0), F(3, 0)
dV(1, 0), F(2, 0)

2Find the center and radius of the circle whose equation is x2 + 10x + y2 = 75.
aC(�10, 0), r = 100
bC(�10, 0), r = 10
cC(�5, 0), r = 100
dC(�5, 0), r = 10

3Find the foci of the ellipse with the following equation.
((x-1)�2�9)+((Y+2)�2�25)=1
aF1(5, �2), F2(�3, �2)
bF1(1, 2), F2(1, �6)
cF1(4, �2), F2(�2, �2)
dF1(1, 1), F2(1, �5)

4Find the slopes of the asymptotes of a hyperbola with the following equation.
(Y�2�81)-(x�2�64)=1
a8/9
b9/8
c8/9, �8/9
d9/8, �9/8
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
1Find the vertex and focus of the parabola whose equation is 4y = x2 + 4.
2Find the center and radius of the circle whose equation is x2 + 10x + y2 = 75.
3Find the foci of the ellipse with the following equation.
((x-1)�2�9)+((Y+2)�2�25)=1
4Find the slopes of the asymptotes of a hyperbola with the following equation.
(Y�2�81)-(x�2�64)=1
***
1) Find the vertex and focus of the parabola whose equation is 4y = x^2 + 4.
y=x^2/4+4/4
y=x^2/4+1
y=(1/4)(x^2)+1
x^2/4=y-1
x^2=4(y-1)
This is an equation of a parabola of standard form: (x-h)^2=4p(y-k), (h, k) being the (x, y) coordinates of the vertex. Parabola has a vertical axis of symmetry and it opens upwards.
For given equation:
vertex: (0,1)
..
4p=4
p=1
focus: (0,2)
ans:bV(0, 1), F(0, 2)
..
2) Find the center and radius of the circle whose equation is x2 + 10x + y2 = 75.
x^2+10x+y^2=75
complete the square
(x^2+10x+25)+(y^2)=75+25=100
(x+5)^2+(y+0)^2=100
center:(-5,0)
radius=√100=10
ans:dC(�5, 0), r = 10
..
3) Find the foci of the ellipse with the following equation.
((x-1)�2/9)+((Y+2)�2/25)=1
This is an equation of an ellipse with vertical major axis of the standard form:
(x-h)^2/b^2+(y-k)^2/a^2=1, a>b, with (h,k) being the (x,y) coordinates of the center.
center: (1,-2)
a^2=25
b^2=9
c^2=a^2-b^2=25-9=16
c=√16=4
Foci: (1,-2�c)=(1,-2�4)=(1,2) and (1,-6)
ans: bF1(1, 2), F2(1, �6)
..
4Find the slopes of the asymptotes of a hyperbola with the following equation.
(Y�2/81)-(x�2/64)=1
This is an equation of a hyperbola with vertical transverse axis of the standard form:
(y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center
For given equation:
Center: (0,0)
a^2=81
a=√81=9
..
b^2=64
b=√64=8
slope of asymptotes=�a/b=�9/8
ans:
d9/8, �9/8
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