Given one focus of an ellipse, two tangent lines, and the point of tangency with one of those lines, construct the other focus.

Question

Focus $F'$ of the ellipse, point $D_1$ and two tangents $t_1$ and $t_2$ are given. Let $D_1$ be the touching point of the tangent $t_1$ and the ellipse. Construct the second focus of the ellipse.

I found the axisymmetric images of point $F'$ with respect to tangents $t_1$ and $t_2$, $G_1$ and $G_2$. Then I found the bisector $h$ of length $G_1G_2$. And now I know that the second focus of the ellipse is somewhere on the line $h$. But I need one more condition to determine the exact position of the second focus.

Answer 1

(interactive version here : moving the blue point gives a variable tangent ; changing the value of $b$ controls the eccentricity of the ellipse ; the choice has been made to keep $a=1$ fixed, giving $f=\sqrt{1-b^2}$). The radius of the circle is $2a=2$.

Three steps :

• The set of symmetrical (or "mirror") points of focus $F'$ with respect to all tangent lines is known to be a circle centered in the other focus $F$.

• Therefore, a fact you have remarked, having two of these mirror points, $G_1$ and $G_2$, $G_1G_2$ is a chord of this circle ; as a consequence, the bissector line of this chord passes through the circle's center $F$.

• Tangency point $D_1$ being known, the reflection property of the ellipse implies that line $F_1D_1$ passes through focus $F$.

Conclusion : $F$ is defined as the intersection of line $F_1D_1$ with the bissector line of $[G_1G_2]$.

Answer 2

I suspect there has to be used the notion of "focus".

We know two facts: The ellipse is the curve created by a straight rod between the focal points. The rod is a light ray, reflected at any point on the ellipse, because all rays emerging from one focal point meet in the second point. It follows, that there has to found two rays, reflected at the two tangents, that meet on a common point on line h,