SOLUTION: What are the intersection points (at least three ordered par solutions) between a circle and parabola with equations X^2 + y^2 + 2x - 4y + 1 = 0 and y^2 - 2x - 4y - 2 = 0 ?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What are the intersection points (at least three ordered par solutions) between a circle and parabola with equations X^2 + y^2 + 2x - 4y + 1 = 0 and y^2 - 2x - 4y - 2 = 0 ?      Log On


   

Question 851372: What are the intersection points (at least three ordered par solutions) between a circle and parabola with equations X^2 + y^2 + 2x - 4y + 1 = 0 and y^2 - 2x - 4y - 2 = 0 ?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B+y%5E2+%2B+2x+-+4y+%2B+1+=+0
x%5E2%2B2x%2B1%2By%5E2-4y%2B4%2B1-1-4=0
1.%28x%2B1%29%5E2%2B%28y-2%29%5E2=4
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.
.
y%5E2-2x-4y-2=0
2x=y%5E2-4y%2B4-2-4
2.2x=%28y-2%29%5E2-6
.
.
.
From eq. 2,
%28y-2%29%5E2=2x%2B6
Substitute into eq. 1,
%28x%2B1%29%5E2%2B%282x%2B6%29=4
x%5E2%2B2x%2B1%2B2x%2B6=4
x%5E2%2B4x%2B3=0
%28x%2B1%29%28x%2B3%29=0
x%2B1=0
x=-1
Then
%28y-2%29%5E2=2%28-1%29%2B6
%28y-2%29%5E2=4
y-2=0+%2B-+2
y=0 and y=4
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.
.
x%2B3=0
x=-3
Then,
%28y-2%29%5E2=2%28-3%29%2B6
y-2=0
y=2
So then the intersection points are (-1,0), (-1,4), (-3,2)