Reading through a handout?, I noticed a section (section 12.2.3, page 308) that stated you can construct tangents from an external point to an ellipse by noticing that "the circle with diameter PF is tangent to the auxiliary circle at two points on the tangent to the ellipse at P." (the exact wording is different but I think this is what the author meant).
Surprisingly, unlike other statements in the handout, this one is not followed by a proof. How would one go about synthetically proving this theorem/construction? I thought a length or maybe collinearity approach would work (because other constructions of tangent lines like Fermat's uses length ratios), but am not sure how to proceed. As a last resort, maybe I need to use coordinate geometry, as that is how the previous construction "Construction of tangent at P (on the ellipse)" is proven. However, unlike that one, the P in this scenario is arbitrary and so it would be hard to set up a nice coordinate proof.
Any help would be much appreciated!
Best Answer
This construction derives from another construction, which I'll repeat here for the ease of the reader.
If $PX$ is a tangent from $P$ to the ellipse and ${F'}_1$ is the reflection of $F_1$ about $PX$, then ${F'}_1XF_2$ are aligned (by the reflection property of the ellipse), $F_2{F'}_1=2a$ (major axis of the ellipse) and $PX$ is the bisector of $\angle F_1P{F'}_1$ (see figure below). Hence point ${F'}_1$ can be constructed as an intersection of the circle of center $F_2$ and radius $2a$ with circle of center $P$ and radius $PF_1$. Once ${F'}_1$ has been found, tangent $PX$ can be constructed. Needless to say, the other intersection ${F''}_1$ gives the other tangent.
Consider now the midpoint $M$ of $F_1{F'}_1$, which is a point on tangent $PX$. It lies on the circle of diameter $PF_1$, because $\angle PMF_1=90°$. And it lies on the auxiliary circle of the ellipse, because $OM={1\over2}F_2{F'}_1=a$. Hence, we find the nice result we wanted to prove.