How can I determine coordinates of vertex of parabola as intersection of a plane and a cone?

I want to determine the point I as vertex of the parabola that is the intersection of the plane plane = InfinitePlane[{{0, myy, 0}, {0, myy, -1}, {1, myy, -3}}]; and the cone = Cone[{pO, pA}, myr]; My code

myy = -3;
myh = 5;
myr = 5;
pA = {0, 0, myr};
pO = {0, 0, 0};
cone = Cone[{pO, pA}, myr];
plane = InfinitePlane[{{0, myy, 0}, {0, myy, -1}, {1, myy, -3}}];
Show[Graphics3D[{Opacity[0.8], cone, plane,
Text[Style["A", Italic, 14, FontFamily -> "Times"], {0, 0,
myr + 0.3}],
Text[Style["O", Italic, 14, FontFamily -> "Times"], {0, 0, -0.4}],
Thick, Dashed, Red, Line[{{pO, pA}}]}, Boxed -> False]]

You can use Reduce to find intersection:

i = Reduce[{x, y, z} \[Element] cone && {x, y, z} \[Element]
plane, {x, y, z}]
p = ParametricPlot3D[{t, -3, 5 - Sqrt[9 + t^2]}, {t, -4, 4},
PlotStyle -> Red]
g = Show[
Graphics3D[{Opacity[0.8], cone, plane,
Text[Style["A", Italic, 14, FontFamily -> "Times"], {0, 0,
myr + 0.3}],
Text[Style["O", Italic, 14, FontFamily -> "Times"], {0, 0, -0.4}],
Thick, Dashed, Red, Line[{{pO, pA}}]}, Boxed -> False], p]

(x == -4 && y == -3 && z == 0) || (-4 < x < 4 && y == -3 && 0 <= z <= 5 - Sqrt[9 + x^2]) || (x == 4 && y == -3 && z == 0)

The middle solution was plotted:

Need to contain the intersection line of the bottom of the cone?

plot3d =
RegionPlot3D[DiscretizeRegion@RegionIntersection[cone, plane],
BoundaryStyle -> {Thick, Red}, PlotStyle -> None];
Show[plot3d,
Graphics3D[{Opacity[0.8], cone, plane,
Text[Style["A", Italic, 14, FontFamily -> "Times"], {0, 0,
myr + 0.3}],
Text[Style["O", Italic, 14, FontFamily -> "Times"], {0, 0, -0.4}],
Thick, Dashed, Red, Line[{{pO, pA}}]}, Boxed -> False]]

• I need to contain the intersection line of the bottom of the cone. Commented Mar 24 at 5:31