Find intersections points with axes

Question

Let's create some points corresponding to a simple closed curve

C0 = ContourPlot[x^2/4 + y^2/9 == 1, {x, -5, 5}, {y, -5, 5}, PlotPoints -> 100];
data = C0[[1, 1]];
S0 = ListPlot[data, PlotStyle -> {Blue, PointSize[0.005]}]

Now I would like to find the intersection points with the two axes. In this example the solutions should be $x = \pm 2$ and $y = \pm 3$.

IMPORTANT NOTE: The real data file corresponds to a closed curve with unknown analytical equation, so in the suggested solution you should not take into account the equation of the ellipse, only data is known.

Many thanks in advance.

Answer 1

Get rid of duplicate x values we can feed to an interpolation function

domain = Select[Tally[data[[All, 1]]], #[[2]] == 1 &][[All, 1]];

Create Interpolation functions and evaluate at 0 These are for the top and bottom halves of the fn

Interpolation[Select[data, #[[2]] < 0 && MemberQ[domain, #[[1]]] &]][0]
Interpolation[Select[data, #[[2]] > 0 && MemberQ[domain, #[[1]]] &]][0]

-2.99994
2.99994

Max and Min on the domain can approximate the x intercepts:

Max[domain]
Min[domain]

1.99985
-1.99985

EDIT

Assuming your ellipse isn't 0-centered you will need exact x-intercepts

This requires a second domain and second set of interpolating functions, necessitating a transposition and reevaluation.

data2 = Reverse /@ data;
domain2 = Select[Tally[data2[[All, 1]]], #[[2]] == 1 &][[All, 1]];


Interpolation[Select[data2, #[[2]] < 0 && MemberQ[domain2, #[[1]]] &]][0]
Interpolation[Select[data2, #[[2]] > 0 && MemberQ[domain2, #[[1]]] &]][0]

-1.99996
1.99996