How to get exact point of intersection of a plot with axis?

Question

I have a parametric plot y(t) vs. x(t). I want to know exactly at what value of x, this y vs x curve crosses the x-axis? I tried with "Get coordinates" of Drawing tools but this is not so accurate. Can anyone please tell me how I can print the exact point of intersection between this curve and x-axis?

I want to find at what value of t, dbdt becomes zero.

Answer 1

You may use MeshFunctions, though I don't know the precision :

gr00 = ParametricPlot[{t, t^2 - 1}, {t, 0, 2}, Mesh -> {{0}}, 
  MeshFunctions ->  {#2 &}, 
  MeshStyle -> Directive[Black, PointSize[.05]]]

Cases[Normal[gr00], Point[___], {1, Infinity}]  

{Point[{0.999811, 0.}]}

Answer 2

Consider

x[u_] := Sin[u]
y[u_] := Sin[2 u]
ParametricPlot[{x[u], y[u]}, {u, 0, 2 Pi}]

Solve for u an equation $y(u)=0$:

sol = u /. Solve[y[u] == 0, u]

{ConditionalExpression[[Pi] C1, C1 [Element] Integers], ConditionalExpression[1/2 ([Pi] + 2 [Pi] C1), C1 [Element] Integers]}

Evaluate x[u] taking sol as input and taking into account the conditions from the ConditionalExpressions:

Table[Simplify[x[sol[[i]]], Assumptions -> sol[[i, 2]]], {i, 1, Length@sol}]

{0, (-1)^C1}

One probably has to check what values can C[1] take (in this case Integers), so maybe instead

Simplify[x[sol], Assumptions -> sol]

{ConditionalExpression[Sin[[Pi] C1], C1 [Element] Integers], ConditionalExpression[Cos[[Pi] C1], C1 [Element] Integers]}

which on the other hand returns a slightly less clear solution.

---

EDIT: I developed a different method, suitable for this problem also, while answering another question.

Let's take the plot in the form

x[u_] := Sin[u]
y[u_] := Sin[2 u]
curve = Show[
  ParametricPlot[{x[u], y[u]}, {u, 0, 2 Pi}, Axes -> None, 
   PlotRangePadding -> None], 
  Plot[0, {x, -1, 1}, Axes -> None, PlotRangePadding -> None]]

and extract the pixel positions of intersection with

px = PixelValuePositions[#, White] & @
  MorphologicalBranchPoints @ Thinning @ Binarize @ ColorNegate @ curve

{{180, 182}, {2, 180}, {180, 180}, {359, 180}, {180, 179}}

We need to connect them to the actual coordinates on the plot:

pl = PlotRange@curve

{{-1., 1.}, {-1., 0.999999}}

id = ImageDimensions@curve

{360, 360}

The relation between pl and id is linear, $y=ax+b$ and $y=cx+d$ in the horizontal and vertical directions, respecively:

{a, b} = {a, b} /. 
  First@Solve[{pl[[1, 1]] == b, pl[[1, 2]] == a id[[1]] + b}, {a, b}]
{c, d} = {c, d} /. 
  First@Solve[{pl[[2, 1]] == d, pl[[2, 2]] == c id[[2]] + d}, {c, d}]

{0.00555555, -1.}

{0.00555555, -1.}

The pixel positions transformed to plot coordinates:

ic = {a #1 + b, c #2 + d} & @@@ px

{{9.49664*10^-9, 0.011111}, {-0.988889, -1.50139*10^-7}, {9.49664*10^-9, \ -1.50139*10^-7}, {0.994444, -1.50139*10^-7}, {9.49664*10^-9, \ -0.0055557}}

look like this:

We can get rid of the ambiguity (5 points for 3 intersections) with clustering:

clu = ClusterClassify[ic, Method -> "DBSCAN"]

Number of clusters: 3

g = GatherBy[ic, clu]
icmean = Chop[#, 10^-6] &@Reverse[Mean /@ g]

{{0.994444, 0}, {-0.988889, 0}, {0, 0.0018517}}

Show[curve, Graphics[{PointSize[Large], Point[#]}] &@icmean]

which looks very good; icmean are the positions of the three intersections.