Answer Page
 
 
 
1) 
 
 
 
2) 
 
 
3) 
 
 
4) 
 
 
 
5) 
 
 
6)  V(2,8) foci at(2, -4), (2, 4) 
 
Ellipse with center at (2, 0)  Half way between the two foci's.
a = 8 , a2 = 64  distance from center to vertex
c = 4, c2 = 16  distance from center to focus.
b2 = 48  (a2 - c2)
 
(x - 2)2  +  y2
                                                                   --------     ----     =   1
                                                                     48         64
 
 
 
7)  V(3,1)  F(3, 3) 
 
p = 2  Distance from vertex to focus.  Parabola opens up.
4p = 8
 
y - 1 = (1/8)(x - 3)2
 
 
 
8)  C(2, -3)  V(7, -3)  E(2, 1)
 
The transverse axis is horizontal with:
a = 5, a2 = 25  (distance from center to vertex)
b = 4, b2 = 16  (distance from center to end of conjugate axis)
 
(x - 2)2       (y + 3)2
                                                              --------    -   ---------   =   1
                                                                25               16
 
 
 
9)
x2 + y2 = 25 and x2 + 10y2 = 169
Solve for x2 in the first equation
1)  x2 = -y2 + 25
Substitute in the second equation
(-y2 + 25) + 10y2 = 169
Combine like terms
9y2 + 25 = 169
Solve for y
9y2 = 144
y2 = 144/9 = 16
y = + 4
Substitute this back into equation 1 above to find the x values
x2 = -(+ 4)2 + 25
x2 = 9
x = + 3
 
There are four intersection points:  ( 3, 4), (3, -4), (-3, 4) and (-3, -4)
 
 
 
10)
y2 - x2 = 64  and x2 + y2 = 25
Isolate for y2 in the first equation
1)  y2 = x2 + 64
Substitute into the second equation
x2 + x2 + 64 = 25
2x2 + 64 = 25
Solve for x
2x2 = - 39
x2 = -39/2
This yields imaginary answers, since we would be taking the square root of a negative number.  This tells us the graphs do not intersect!!
 
That should do it for another chapter!!
It's time to head toward trig!!
 
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