This problem comes from the university entrance exam in 2016. So in the original answer, calculus knowledge is unnecessary. But I am looking for the possibility of any simpler solution when using calculus or differential geometry.
See the figure above. The ellipse in red has fixed minor axis, the equation of which is $$\dfrac{x^2}{a^2}+y^2=1,\quad a>1$$
If the ellipse has at most three points in common with circles centerred at $O(0,1)$, what is the range of $a$ and its eccentricity $e=\dfrac{\sqrt{a^2-1}}{a}$?
The answer: the range of $a$ is $1<a^2<2$, that of $e$: $0<e<\dfrac{\sqrt{2}}{2}$:
From the answer above, we can see that the curvature of the red circle should also be at a specific range compared to the blue circle with radius 2. Is there any simple relationship between the curvature of an ellipse and its eccentricity? Does there exist any simple solution by using the advanced concepts in differential geometry?
You are essentially asking what is the major axis of the largest ellipse, contained in a unit circle, whose minor axis is a radius of such a circle. That can be computed by imposing that $$\left\{\begin{array}{rcl}\frac{x^2}{a^2}+y^2 &=& 1 \\ x^2+(y-1)^2 &=& 4\end{array}\right. $$ has the only solution $(x,y)=(0,-1)$, i.e. by imposing that a discriminant is zero.
As an alternative, we may impose that the osculating circle of the ellipse $\frac{x^2}{a^2}+y^2$ at the point $(0,-1)$ is exactly the circle $x^2+(y-1)^2=4$. That is equivalent to equating two second derivatives or two curvatures. The curvature of an ellipse, obviously, is not constant, but the curvature at the vertices, almost as obviously, just depends on the $a,b$ parameters.
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