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

2 Answers
t0hierry
Jan 25, 2017

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)

Shwetank Mauria
Jan 25, 2017

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]}

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