given a circle of radius R and an ellipse. If all four vertices of the ellipse are contained within the circle, then is the entire ellipse contained within the circle.
Note: points of the ellipse can be on the circle (namely the vertices)
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What do you mean by the vertices of an ellipse? – Michael Albanese Mar 17 '22 at 05:42
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I assume the endpoints of the major and minor axes. – Tbw Mar 17 '22 at 05:43
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Do the circle and ellipse have to share the same center? If not, this is not always true.(picture, graph). – Graviton Mar 17 '22 at 05:51
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If you want to find out if they do intersect given the vertices, the answer is pretty hard but I found this answer and this document (section 4.2 on). You can get more info and more Stack Exchange posts by searching "ellipse intersection test." – Tbw Mar 17 '22 at 06:12
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thanks everyone. After I thought about it, there are lots of ways to show this is false. I looked at radius of curvature at the minor axis vertex. That link is very useful – Matthew Schroeder Mar 17 '22 at 06:17
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Having all four vertices in the circle does not mean that the whole ellipse will be in the circle. One specific counterexample would be $$\frac{3}{2}x^{2}+4\left(y-\frac{1}{2}\right)^{2}=1$$ whose vertices lie inside the unit circle but parts of the ellipse lie outside, as seen in this desmos graph.
Edit: Even a circle can work: $$4\left(x-\frac{2}{5}\right)^2+4\left(y-\frac{2}{5}\right)^{2}=1$$
Tbw
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one of the vertices in your example is on the circle, which OP has prohibited – Graviton Mar 17 '22 at 05:52
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He stated "Note: points of the ellipse can be on the circle (namely the vertices)" – Tbw Mar 17 '22 at 05:54
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1And $-1/2$ can be replaced by $-1/2+.01$ if you want to make the vertex not touch. – Tbw Mar 17 '22 at 05:56