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Solve the system of equations (conics)

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)

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