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

2 Answers
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)

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]}