Given an ellipse, we want to draw a set of tangents to it, which are orthogonal to each other. We are looking for a synthetic way; not through algebra or trigonometry.

I know how to construct one tangent: Given any point $P$ on the ellipse, we connect it with the two foci, say $F_1$ and $F_2$ and then we bisect angle $F_1PF_2$. The bisector is the "normal". Then we draw an orthogonal line to the bisector, from point $P$f and this is our tangent.

I am trying to think of a way to construct another tangent, orthogonal to it. Clearly the normal of the second tangent will be orthogonal to the 1st normal. We are looking for the single point $Q$ (or maybe 2 points), for which, the 2nd normal will bisect angle $F_1QF_2$ but I can't think of any way to construct it.

Any ideas? Thank you!

  • $\begingroup$ Sorry but "which are vertical to each other" doesn't mean anything... Please correct it. Maybe you wanted to say "orthogonal" ? $\endgroup$
    – Jean Marie
    Jul 5, 2021 at 10:07
  • $\begingroup$ en.wikipedia.org/wiki/Director_circle $\endgroup$ Jul 5, 2021 at 10:41
  • $\begingroup$ @JeanMarie sorry but English is not my mother tongue. I corrected it. $\endgroup$ Jul 5, 2021 at 11:00
  • $\begingroup$ I have changed in your question all remaining adjectives "parallel" into "orthogonal" which is the most appropriate term for 2 lines making a 90° angle. $\endgroup$
    – Jean Marie
    Jul 5, 2021 at 11:34

2 Answers 2


There is a simple way to construct a second tangent, perpendicular to the first one you constructed at $P$. Let the normal at $P$ intersect again the ellipse at $N$ and let $M$ be the midpoint of $PN$. The line passing through $M$ and the center $O$ of the ellipse will then intersect the ellipse at two points $Q$ and $Q'$, with the property that the tangents at both points are parallel to $PN$, and hence perpendicular to the first tangent.

As I wrote in a comment, the intersections between the tangents at $Q$ and $Q'$ and the tangent at $P$ lie on the director circle of the ellipse.

enter image description here

  • $\begingroup$ Brilliant solution! Is there a proof? $\endgroup$ Jul 6, 2021 at 11:47
  • $\begingroup$ If the Wikipedia proof I linked is not enough, I could try to provide another proof. $\endgroup$ Jul 6, 2021 at 13:00

This being my first contribution to the forum, let me hope that I got the formatting and importing procedures correct.

I have constructed a tangent through point P just as you said, bisecting ∠FPF₂, and constructing the tangent perpendicular to that bisector.

Through center O I have constructed a diameter perpendicular to the tangent line at P. It meets the ellipse at Q, and there I have constructed another tangent line using the same angle bisection method.

Through O I have constructed a second diameter, parallel to the tangent at Q. This diameter meets the ellipse at points A and B.

Lines OQ and AOB, are conjugate diameters. I have constructed lines through A and B, both parallel to OQ, which makes them tangent to the ellipse. It also makes them perpendicular to the tangent through P.

ellipse construction


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