Find an ellipse tangent to triangle given centre position of the ellipse

Question

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?

Answer 1

I like the more-geometric answers, but I'm somewhat committed to this analytic approach, so let's continue ...

Place $O$ at the origin, and let $P = (-d,0)$. As @Aretino helpfully observes in a comment, $\overleftrightarrow{OD}$ must bisect the chord $\overline{T_1T_2}$, so define $M$ as midpoint of the chord, with $|PM|=m$, so that $M=(-d+m,0)$; also, define $\theta = \angle OMT_1 = \beta+\delta$. Then, with $a := |T_1T_2|$, we can write $$T_1 = M + \frac{a}2(\cos\theta,\sin\theta) \qquad T_2 = M - \frac{a}2(\cos\theta,\sin\theta) \tag{1}$$

Since $O$ is the center of the ellipse, we can define, say, $T_3=-T_1$ to get a third point on the ellipse. Since $T_1$ and $T_2$ are points of tangency, we can tease them into double-points by defining $$T_1' = T_1 + t_1(P-T_1) \qquad T_2' = T_2 + t_2(P-T_2) \tag{2}$$ for "infinitesimally small" $t_1$ and $t_2$ that we can treat as non-zero or zero as to our advantage.

With five points on the conic, we can use a determinant to get its equation. (See, for instance, this answer.) Using a computer algebra system such as Mathematica to expand the determinant, we obtain factors of $t_1$ and $t_2$ that we divide-out (because they're non-zero), then set the remaining instances of these values to $0$ (because they're not non-zero), so that the equation becomes ...

$$\begin{align} 0 &= x^2 a^2 \sin^2\theta - 2 a^2 x y \cos\theta \sin\theta + y^2\left( 4 m (d-m) + a^2 \cos^2\theta \right) \\ &- a^2d\sin^2\theta \left( d - m \right) \end{align}\tag{3}$$

From here, we can consult, say, this answer for recipes that express metric properties of a conic in terms of the coefficients of the general second-degree polynomial. We find that the angle $\phi$ that the major axis makes with the $x$ axis satisfies

$$\tan2\phi = \frac{a^2 \sin 2\theta}{4 (d - m) m + a^2 \cos 2\theta} \tag{4}$$

and the major and minor radii of the ellipse are given by

$$r_{\pm}^2 = \frac{d}{8m}\left( a^2 + 4m (d - m) \pm \sqrt{ a^4 + 16 m^2 (d - m)^2 + 8 a^2 m (d - m) \cos 2\theta )}\right) \tag{5}$$

To rewrite in terms of the problem's stated parameters, the Law of Cosines gives $$a^2 = b^2 + c^2 - 2 b c\cos\alpha \tag{6}$$ where $\alpha := \angle T_1PT_2 = 180^\circ - \beta - \gamma$. We can use Stewart's Theorem to show $$m^2 = \frac14\left(-a^2+2b^2+2c^2\right) = \frac14\left( b^2 + c^2 + 2b c \cos\alpha\right) \tag{7}$$ Also, since $|\triangle PT_1T_2| = \frac12b c \sin\alpha = \frac12am\sin\theta$, we can find $$\begin{align} \sin^2\theta &= \frac{4b^2 c^2 \sin^2\alpha}{(b^2+c^2-2b c\cos\alpha)(b^2+c^2+2b c \cos\alpha)} = \frac{4b^2c^2\sin^2\alpha}{(b^2-c^2)^2+4b^2c^2\sin^2\alpha} \tag{8} \\[4pt] \cos^2\theta &= \frac{(b^2-c^2)^2}{(b^2-c^2)^2+4b^2c^2\sin^2\alpha} \tag{9} \\[4pt] \cos2\theta &= \frac{(b^2-c^2)^2-4b^2c^2\sin^2\alpha}{(b^2-c^2)^2+4b^2c^2\sin^2\alpha} \tag{10} \end{align}$$ Expressions $(4)$ and $(5)$ don't seem to get appreciably better by substituting-in from these expressions and simplifying (the arbitrary $d$ gets in the way), so I'll leave things here. $\square$

Answer 2

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.

Answer 3

After construction of the third tangent point $Q$ as described in the other answer, we can consider the construction as the triangle $ABC$ with generalized Steiner inellipse, for which the foci can be found as the roots of derivative of the function \begin{align} p(z)&=(z-A)^u(z-B)^v(z-C)^w ,\quad u,v,w>0,\ u+v+w=1 . \end{align}

In this case

\begin{align} u&= \frac{|AT_2|\cdot|CQ|}{|CT_2|\cdot|BC|+|AT_2|\cdot|CQ|} ,\\ v&= \frac{|CT_2|\cdot|BQ|}{|CT_2|\cdot|BC|+|AT_2|\cdot|CQ|} ,\\ w&=1-u-v \end{align}

and the foci are located at

\begin{align} F_{1,2}&= \tfrac12\,(u+v+w)^{-1}\cdot \Big( (v+w)\,A+ (w+u)\,B+ (u+v)\,C \\ &\pm \left( (w+v)^2\,A^2 +(u+w)^2\,B^2 +(u+v)^2\,C^2 \right. \\ &+2\,(u\,v-v\,w-w\,u-w^2)\,A\cdot B \\ &+2\,(v\,w-v\,u-w\,u-u^2)\,B\cdot C \\ &+\left. 2\,(w\,u-v\,w-v\,u-v^2)\,C\cdot A \right)^{1/2} \Big) . \end{align}