Touching point of ellipses

Question

In celestial mechanics the following problem from elementary geometry arises.

Let two ellipses with one common focus $o=(0,0)$ be given such that the smaller (black) touches the larger (blue), see image below. We are given

• the minimal $R_1$ and maximal distance $R_2$ to the focus for the larger ellipse
• the location of the point $q$ on the smaller ellipse with minimal distance $q_1$ to the focus
• the angle $\theta$ between their semi-major axes

What is the distance $R$ of the touching point $p$ from the focus?

Any tips how to tackle this problem, thanks in advance!

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Update:

Using the ellipse polar form equation for the large ellipse $$R=\frac{a (1-e^2)}{1+e \cos \beta}=\frac{2 R_1 R_2}{ (R_2 - R_1)\cos \beta + R_1 + R_2},\tag{1}$$ where $\beta$ is the angle between $R_1$ and $R$. Applying the same formula for the smaller ellipse gives $$R=\frac{2 q_1 q_2}{ (q_2 - q_1)\cos (\beta - \theta ) + q_1 + q_2}.\tag{2}$$ Combined we get \begin{align*} R_1 R_2 \left[(q_2 - q_1)\cos (\beta - \theta ) + q_1 + q_2\right] = q_1 q_2 \left[(R_2 - R_1)\cos (\beta) + R_1 + R_2\right].\tag{3} \end{align*} As both are touching we can differentiate in $\beta$ to get \begin{align*} R_1 R_2 (q_2 - q_1)\sin (\beta - \theta ) = q_1 q_2 (R_2 - R_1)\sin (\beta).\tag{4} \end{align*} Rearranging (3) and (4) as formulas for $q_2$ gives \begin{align*} \tfrac{q_1 R_1 R_2(-\cos (\beta - \theta )+1 )}{q_1 (R_2 - R_1)\cos (\beta) + q_1(R_1 + R_2) - R_1 R_2 (\cos (\beta - \theta )+1)} = \tfrac{q_1 R_1 R_2\sin (\beta - \theta )}{R_1 R_2\sin (\beta - \theta ) - q_1 (R_2 - R_1)\sin (\beta)} \end{align*} or \begin{align*} \frac{2 R_1 R_2 - q_1 (R_1+R_2) }{q_1 (R_2-R_1)}= \cos(\beta)+\sin(\beta) \tan\tfrac{\beta-\theta}{2} .\tag{5} \end{align*} The only solution is given by \begin{align*} \cos\beta= \frac{2 c-(1+c^2) \cos\theta}{1+c^2-2 c \cos \theta}. \end{align*} where $c$ is the term on the left side of (5). In combination with (1) we get $R$.

Answer 1

Large ellipse in polar coordinates $(\phi,\rho)$:

$$\rho=\dfrac{2R_1R_2}{R_2+R_1+(R_2-R_1)\cos(\phi)}$$

Small ellipse in polar coordinates:

$$\rho=\dfrac{2q_1q_2}{q_2+q_1+(q_2-q_1)\cos(\phi-\theta)}.$$

Equating the two we express the common point(s) between the ellipses:

$$2q_1q_2(R_2+R_1+(R_2-R_1)\cos(\phi))=2R_1R_2(q_2+q_1+(q_2-q_1)\cos(\phi-\theta)).$$

The tangency condition is obtained by differentiating on $\phi$:

$$2q_1q_2(R_2-R_1)\sin(\phi)=2R_1R_2(q_2-q_1)\sin(\phi-\theta).$$

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Update:

After replacing the $R$ and the $q$ by their inverses and eliminating $q_2-q_1$, we get

$$(R_2-R_1)(\sin(2\phi-\theta)+\sin(\phi-\theta)-\sin(\phi))=2q_1\sin(\phi-\theta).$$

Calculation errors not excluded.

Answer 2

Let $A$, $B$ be the foci of the large ellipse, with given major axis $CH=2a$ and distance between foci $AB=2c$ (see figure below - sorry but I didn't use your notation).

Vertex $D$ of the small ellipse (with foci $A$ and $E$) is given, hence we know $p=AD$ and $\theta=\angle EAB$. I want to find $AE=x$, knowing that the ellipses are tangent at $F$.

From the cosine law we get: $$ BE^2=x^2+4c^2-4cx\cos\theta $$

and we know that

$$ DG=2p+x=AF+EF=(2a-BF)+(BF-BE)=2a-BE, $$ that is: $$ BE=2a-2p-x. $$

We thus get the equation $$x^2+4c^2-4cx\cos\theta=(2a-2p-x)^2$$ which is a linear equation for $x$, giving: $$ AE=x={a^2+p^2-c^2-2ap\over a-p-c\cos\theta}. $$

Once $AE$ is found, one can also compute $AF$. In fact, if $\beta=\angle ABF$ then by the cosine rule in triangles $ABE$ and $ABF$ we get:

$$ \cos\beta={4c^2+BE^2-AE^2\over 4c\cdot BE} =\frac{a^2-a (2 p+x)+c^2+p (p+x)}{c (2 a-2 p-x)} $$ and $$ AF={a^2+c^2-2ac\cos\beta\over a-c \cos\beta} =\frac{(2p+x)(a^2-c^2)-2 a p (p+x)}{a^2-c^2-p (p+x)}. $$