Bsrgvty

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]

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]]]

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]

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 == frac5-4x^2 - x + x^2/42)$

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.