Q: Solve the system of equations (both of which are conic sections)
1: x2 + y2 – 20x + 8y + 7 = 0
2: 9x2 + y2 + 4x + 8y + 7 = 0
A: I am going to solve this by using the elimination method (since I see that I can cancel out all of the y’s)
I am going to multiply equation 1 by (-1) and then add it to equation 2:
2: 9x2 + y2 + 4x + 8y + 7 = 0
1: -x2 – y2 + 20x – 8y – 7 = 0 [after it has been multiplied by -1]
Now, add them together to get equation 3:
3: 8x2 +24x = 0
Now, equation 3 only has x’s so we can solve for x (by factoring):
(8x)(x + 3) = 0
So, x = 0 or -3
But, we aren’t done… Find the solution to this system of conics means that we are finding the point(s) of intersection — and it turns out there are two points of intersection since there were two solutions for x. So, our points (solutions) are:
(0, ?) and (-3, ?)
We need to find the y-coordinate of each solution separately. We do this by plugging in the x value to either of the original equations (both will lead us to the same correct solution):
First x-value: (0, ?)
x2 + y2 – 20x + 8y + 7 = 0
(0)2 + y2 – 20(0) + 8y + 7 = 0
y2 + 8y + 7 = 0
Solve for y by factoring:
(y + 7)(y + 1)=0
y = -7, -1
OH! So this really means that there are more than two solutions…. When x = 0, we found two answers for y.
(0, -7) and (0, -1)
Second x-value: (-3, ?)
x2 + y2 – 20x + 8y + 7 = 0
(-3)2 + y2 – 20(-3) + 8y + 7 = 0
9 + y2 +60 + 8y + 7 = 0
y2 + 8y + 76 = 0
Since this does not factor, we need to solve for y by using the quadratic formula. I did not show my work here, but we end up with a negative under the square root, so there are no real solutions for when x = -3.
Therefore, there are two real solutions to this equation (2 points of intersection) and they are:
(0, -7) and (0, -1)
Here is a visual of what we are solving for….. We had the equations for the two graphed conic sections and we found the two points of intersection shown below: