How do you solve the system #x^2+4y^2-4x-8y+4=0# and #x^2+4y-4=0#?
2 Answers
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)
Solutions are
Explanation:
Solving the systems of equations
and
means identifying set of values of
from (B), we get
or
or
or
or
or
or
As the discriminant for
Now when
Hence solutions are
graph{(x^2+4y^2-4x-8y+4)(x^2+4y-4)=0 [-5, 5, -2.5, 2.5]}