Prove that if a circle centred at the focus of a confocal ellipse and hyperbola touches the ellipse internally, it touches the hyperbola at its vertex.

This question can be solved by the method I've mentioned below, but I'm looking for a shorter method.

Let us assume a hyperbola centred at $(0,0$) with foci, $(x_1,0)$ and $(-x_1,0)$ and semi major axes length $a_1$. The equation of the hyperbola is $$\left(\sqrt{(x-x_1)^2+y^2}-\sqrt{(x+x_1)^2+y^2}\right)^2=4a_1^2$$

Using the property that difference between the distance of any point on a hyperbola from its foci is equal to two times its semi major axis.

Similarly equation of ellipse with semi major axis $a_2$ centred at $(0,0)$ with foci, $(x_1,0)$ and $(-x_1,0)$ is $$\left(\sqrt{(x-x_1)^2+y^2}+\sqrt{(x+x_1)^2+y^2}\right)^2=4a_2^2$$

To proceed I need to find the radius of the circle. The circle will be tangent to the ellipse at the point which is nearest from the foci. Let us take the focus as $(a_1e,0)$, where e is the eccentricity of the ellipse. That point will be $(a_1\cos(\theta)+b_1\sin(\theta))$ where $\theta$ is a parameter and $b_1$ is the length of the semi minor axis. So we need to find the minimum value of the equation:


After we prove the circle touches the ellipse at the end points of the major axis, we just need to prove the centre of the circle is the midpoint between the vertex and the point $(a_1,0)$

I don't know how to proceed further without getting into a very lengthy calculation. Is there a shorter/better method?

  • 1
    $\begingroup$ Use the reflection property of the ellipse to get the point of tangency must be on the $x$-axis. $\endgroup$ Sep 29, 2021 at 14:46
  • $\begingroup$ @user10354138, I can't understand. Could you please elaborate $\endgroup$
    – Tatai
    Sep 29, 2021 at 16:51

1 Answer 1


That property, as written, is clearly false:

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