How can I solve for the intersection points of two ellipses?

Question

I need to find the points at which the two ellipses implicitly defined by

$\qquad y^2=4-4\,x^2 \quad[E1]$

and

$\qquad (1-(x/2))^2+(y-1)^2=1 \quad[E2]$

intersect.

So I isolated $y$ in E2 and then squared it so that I could eliminate it using E1 and then solve for $x$. This turned to be very gnarly.

Is there an easier way of finding the intersection points of E1 and E2? (as shown here in graph)

ContourPlot[
  {(y^2 == 4 - 4*x^2), ((1 - x/2)^2 + (y - 1)^2 == 1)}, {x, -5, 5}, {y, -5, 5}, 
  Frame -> False, Axes -> True]

Answer 1

Solve can be used directly

pts = {x, y} /. 
   Solve[{y^2 == 4 - 4 x^2, (1 - (x/2))^2 + (y - 1)^2 == 1}, {x, y}, 
    Reals] // FullSimplify

(* {{Root[144 - 160*#1 - 328*#1^2 + 
           120*#1^3 + 225*#1^4 & , 1, 
       0], Root[144 - 1280*#1 + 
           1688*#1^2 - 960*#1^3 + 
           225*#1^4 & , 2, 0]}, 
   {Root[144 - 160*#1 - 328*#1^2 + 
           120*#1^3 + 225*#1^4 & , 2, 
       0], Root[144 - 1280*#1 + 
           1688*#1^2 - 960*#1^3 + 
           225*#1^4 & , 1, 0]}} *)

Converting the Root objects to their numeric values

pts // N

(* {{0.539936, 1.68341}, {0.997732, 0.13463}} *)

Show[
 ContourPlot[{
   (y^2 == 4 - 4*x^2),
   ((1 - x/2)^2 + (y - 1)^2 == 1)},
  {x, -1.5, 4.5}, {y, -3, 3},
  Frame -> False,
  Axes -> True],
 Graphics[{Red, AbsolutePointSize[4], Point[pts]}]]

Answer 2

You can (1) use Cases to extract the Lines from ContourPlot output, and (2) use RegionIntersection to find the intersections:

cp = ContourPlot[{(y^2 == 4 - 4*x^2), ((1 - x/2)^2 + (y - 1)^2 == 
      1)}, {x, -5, 5}, {y, -5, 5}, Frame -> False, Axes -> True];

intersections = RegionIntersection @@ Cases[Normal @ cp, _Line, All]

Point[{{0.992916, 0.140285}, {0.539984, 1.68261}}]

Show[cp, Epilog->{Red, PointSize[Large], intersections}]

Show[cp, ListPlot[Callout[#, {##}, Automatic, 
     LabelStyle -> 13, Appearance -> "Frame", LeaderSize -> 30, CalloutStyle -> Red, 
     CalloutMarker -> "BoxPoint"] & /@ intersections[[1]]],
   PlotRange -> {{-3, 4}, {-3, 3}}]

Answer 3

Clearly the easy way is to get a computer program to solve it for you.

If for some reason you want to do it by hand, you could do this:

$(1 - x/2)^2 + (y - 1)^2 == 1$

$(y - 1)^2 == 1-(1 - x/2)^2 == x - x^2/4 $

$y^2 - (y - 1)^2 == 4-4x^2 - (x - x^2/4 )$

$y^2 - (y^2 -2y+ 1) == 4-4x^2 - x + x^2/4 $

$ 2y-1 == 4-4x^2 - x + x^2/4)$

$ y == \frac{5-4x^2 - x + x^2/4}{2})$

And now it's practically a straight binomial.

But I notice this is for Mathematica, and you didn't show why you had a problem using Mathematica. It looked like you were doing stuff by hand.