2
$\begingroup$

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.

enter image description here

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

enter image description hereenter image description here

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?

2 Answers 2

4

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.

$\hspace1.6in$enter image description here

  • 0
    Thank you. (1)Is there any discriminant for such polynomial systems to determine the number of their real solutions? or I will have to convert it into a single variate euqation first?(2)The curvature difference is visually obvious; but as you said, curvature of the ellipse is neither a constant nor they have the same curvature center as the circle. So it is still hard for such intuition to be used as a **proof**. – 2017-01-13
  • 1
    @user6043040: by *discriminant* I mean the discriminant of the second-degree equation you get by contracting that system in a unique equation (by eliminating $x$, for instance). About the second part, there is no intuition to misteriously use, just a curvature to compute at a vertex of an ellipse, through the very definition of the curvature through derivatives. – 2017-01-13
2

SecondOrderContact

Ellipse should make second order contact with osculating circle of radius 1 as shown. $a$ cannot be more than this critical value, else there will be two extra intersections outside red circle.

The evolute of an ellipse can be shown(but not shown), cusped at circle center. The extreme radii of curvature are $a^2/b, b^2/a$ on $y-,x-$ axes, by radius of curvature calculations.

In our case

$$ \frac{a^2}{b} =\frac{a^2}{1}<2 \quad, a < \sqrt2 . \,$$