Find an ellipse tangent to triangle given centre position of the ellipse I have a fully defined triangle, $PT_1T_2$, of side lengths $a$, $b$, & $c$ with corresponding opposite angles $A$, $B$, & $C$. I want to find an ellipse that is tangent to the lines $PT_1$ and $PT_2$ at points $T_1$ and $T_2$ respectively.
As there are many ellipses that satisfy this condition, I impose that the centre point, $O$, of the ellipse is at a distance $d$ from point $P$ and that the line $OP$ is at an angle $D$ from the line $PT_1$, where $0<D<A$.
Given these conditions, can I find the focus of the ellipse and the angle of the major axis relative to the line $PO$? What are the equations for these?


If tangents at $T_1$ and $T_2$ meet at $P$, and $M$ is the midpoint of $T_1T_2$, then line $PM$ passes through the centre $O$ of the ellipse.
Hence in our case centre $O$ must lie on the given line $PM$.
The ellipse is determined when five of its points are known, but once $O$ is fixed symmetric points $T_1'$ and $T_2'$ also belong to the ellipse and we just need another point $Q$.
It is easy to find $Q$ on line $PM$, for in that case the tangent at $Q$ is parallel to $T_1T_2$. If that tangent meets line $PT_1$ at $R$, then $OR$ bisects $T_1Q$ and from Menelaus's theorem we obtain:
$$
OQ=\sqrt{OP\cdot OM}.
$$
From that, point $Q$ and the ellipse can be found.

To find the ellipse one can follow the analytical method outlined by Blue in another answer, but one can also construct its principal axes geometrically.
One can, first of all, find semidiameter $OS$, parallel to $T_1T_2$ and conjugate to $OQ$, from Apollonius's equation:
$$
{OM^2\over OQ^2}+{MT_2^2\over OS^2}=1,
$$
which gives:
$$
OS=MT_2\sqrt{OP\over PM}.
$$
Once semidiameters $OQ$ and $OS$ are known, the axes of the ellipse can be constructed as explained here.
