Question

How do you solve the system `x^2+4y^2-4x-8y+4=0` and `x^2+4y-4=0`?

Answer 1

1/4(x-2)^2 + (y-1)^2 = 1 That is an ellipse. The second equation is a parabola. They both have in common (x=0, y=1) and (x=2, y=0)

Answer 2

Solutions are `(0,1)` and `(2,0)`

Explanation:

Solving the systems of equations

`x^2+4y^2-4x-8y+4=0` ......................(A)

and `x^2+4y-4=0` ......................(B)

means identifying set of values of `x` and `y` which solves both the equations together. In a plane graph, it means the set of points which lie at the intersection of two curves. It is apparent, that while (A) is an ellipse, (B) is a parabola.

from (B), we get `y=-1/4x^2+1` and putting this in (A), we get

`x^2+4(-1/4x^2+1)^2-4x-8(-1/4x^2+1)+4=0`

or `x^2+1/4(-x^2+4)^2-4x+2x^2-8+4=0`

or `4x^2+(x^4-8x^2+16)-16x+8x^2-16=0`

or `x^4+4x^2-16x=0`

or `x(x^3+4x-16)=0`

or `x(x^2(x-2)+2x(x-2)+8(x-2))=0`

or `x(x-2)(x^2+2x+8)=0`

As the discriminant for `x^2+2x+8` is `Delta=2^2--4xx8xx1=-28`, `x^2+2x+8` has no real roots.

Now when `x=0`, `y=1` and when `x=2`, `y=0`

Hence solutions are `(0,1)` and `(2,0)`
graph{(x^2+4y^2-4x-8y+4)(x^2+4y-4)=0 [-5, 5, -2.5, 2.5]}