Constructing the focal points of a conic given a tangent line at a point and the major/minor axes

Now that we have constructed the major and minor axes, we would like to find the focal points of the conic. Since the conic is symmetric across the major/minor axes, then we can construct new points on the conic by reflecting the known points. For example, in the diagram below the point \(A\) is on the conic, and reflecting it across the minor axis will produce a new point \(A'\) on the conic. Again, since everything is symmetric across the minor axis, then we can construct the tangent line at \(A'\) as the reflection of the tangent line at \(A\). The reflective symmetry guarantees that the two tangent lines (coloured green in the diagram below) will intersect at a point \(Q\) on the minor axis.

Now we are in the situation of the Ellipse inscribed in a kite, and we can solve the problem of constructing the focal points in the same way.

Recall the optical property of the ellipse: that the line \(F_1 A\) is the reflection of the line \(F_2 A\) across the tangent line \(XW\). As a consequence, we have \( \angle F_1 A Q = 180^\circ - \angle F_2 A Q\). Since the figure is symmetric across the minor axis of the ellipse, then \(\angle F_2 A Q = \angle F_1 A' Q\), and so we have \(\angle F_1 A Q = 180^\circ - \angle F_1 A' Q\), which is exactly the condition for \(F_1 A Q A'\) to be a cyclic quadrilateral. Therefore \(F_1\) is one of the intersection points of the major axis and the circle through the three points \(A, Q, A'\) (this is the blue circle in the diagram below).

Exactly the same idea (with the roles of \(F_1\) and \(F_2\) reversed) shows that \(F_2\) is the other point of intersection between the major axis and the circle through \(A, Q, A'\).