Find intersections of two ellipses who share one fixed point

Question

Given two ellipses $e_1$ and $e_2$ with $$ e_1 = \{x: \lVert{x - F_1}\rVert + \lVert{x - F_2}\rVert = R \} $$ $$ e_2 = \{ x : \lVert{x - F_1}\rVert + \lVert{x - F_3}\rVert = R \} $$ where $F_1$ is the shared fix point and $R = \lVert{F_3 - F_1}\rVert$. (Note that $R$ is equal for both ellipses)

I know that the intersection points satisfy $\lVert x-F_2\rVert = \lVert x-F_3\rVert$. But I'm stuck here. Is there a simpler way to find the intersection points in this case, than to construct ellipse equations and solve these?

Edit: Based on philipph's answer I constructed the following ideas:

I forgot to mention that my ellipses are neither centered at the origin nor is the major axis along the X-axis. It seemed appropriate to choose the general parametric form for both the ellipses, and the line. First, I looked at the ellipse equation

$$ \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}x_0+a\cos t\cos\alpha-b\sin t\sin\alpha\\y_0+a\cos t\sin\alpha+b\sin t\cos\alpha\end{pmatrix}, \qquad \text{where } 0 \leq t < 2\pi. $$ $\alpha$ is the angle between the X-axis and the major axis, $t$ is the parameter. I constructed for $c_1$ $$ a = \frac{R}{2}=\frac{1}{2}\lVert F_3-F_1\rVert, $$ $$ b_1 = \sqrt{a^2 - \frac{1}{4}\lVert F_1 - F_2\rVert ^2} = \frac{1}{2}\sqrt{\lVert F_3-F_1\rVert^2-\lVert F_1-F_2\rVert ^2}, $$ $$ \begin{pmatrix}x_0\\y_0\end{pmatrix} = \frac{1}{2}\left(F_1+F_2\right) = \frac{1}{2}\begin{pmatrix}(F_1)_1+(F_2)_1\\(F_1)_2+(F_2)_2\end{pmatrix} $$ The final ellipse equation for $c_1$ is $$ \begin{pmatrix}x\\y\end{pmatrix} = \frac{1}{2}\begin{pmatrix} \left( (F_1)_1+(F_2)_1\right)+\lVert F_3-F_1\rVert\cos t\cos\alpha-\sqrt{\lVert F_3-F_1\rVert^2-\lVert F_1-F_2\rVert ^2}\sin t\sin\alpha\\ \left((F_1)_2+(F_2)_2\right)+\lVert F_3-F_1\rVert\cos t\sin\alpha+\sqrt{\lVert F_3-F_1\rVert^2-\lVert F_1-F_2\rVert ^2}\sin t\cos\alpha \end{pmatrix}. $$ The parametric line equation: $$ \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}p_1\\p_2\end{pmatrix}+t\begin{pmatrix}u_1\\u_2\end{pmatrix} $$ and its final form: $$ \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\frac{1}{2}\left((F_2)_1-(F_3)_1\right)\\\frac{1}{2}\left((F_2)_2-(F_3)_2\right)\end{pmatrix} +t\begin{pmatrix}(F_3)_1-(F_2)_1\\(F_2)_2-(F_3)_2\end{pmatrix}. $$ Note that $\vec{p}$ is a vector towards the bisector between $F_2$ and $F_3$, and $\vec{u}$ is its normal. I think the normal should work perfectly in my case?

Now I would try to solve the equations for $t$ and substitute to find the intersection points. Do you think this would work? Did I make any mistake?

Edit 2: I noticed (and actually forgot about before) that $c_2$ is indeed a degenerate ellipse, i.e. a line, because $$ b_2 = \sqrt{a^2 - \frac{1}{4}\lVert F_3 - F_1\rVert^2} = \sqrt{a^2 - a^2} = 0. $$ This yields $$ \begin{pmatrix}x\\y\end{pmatrix} = \frac{1}{2}\begin{pmatrix} \left( (F_1)_1+(F_3)_1\right)+\lVert F_3-F_1\rVert\cos t\cos\alpha\\ \left((F_1)_2+(F_3)_2\right)+\lVert F_3-F_1\rVert\cos t\sin\alpha \end{pmatrix} $$ as ellipse equation for $c_2$. This further simplifies the calculation, and I hope to get some comments on my math and whether or not I made mistakes in the process. Thank you!

Edit 3:

It seems the final equation to get the intersection is rather complicated. I substituted for $t$ and got: $$ \frac{\begin{pmatrix}x\\y\end{pmatrix} - \frac{1}{2}\begin{pmatrix}(F_2)_1-(F_3)_1\\(F_2)_2-(F_3)_2\end{pmatrix}}{\begin{pmatrix}(F_3)_1-(F_2)_1\\(F_2)_2-(F_3)_2\end{pmatrix}} = \arccos\left(\frac{2\begin{pmatrix}x\\y\end{pmatrix}-\begin{pmatrix}(F_1)_1+(F_3)_1\\(F_1)_2+(F_3)_2\end{pmatrix}}{\begin{pmatrix}\lVert F_3-F_1\rVert \cos\alpha\\\lVert F_3-F_1\rVert \cos\alpha\end{pmatrix}}\right) $$ which is sort of like $x = \arccos(x)$ and rather complicated to solve. What did I do wrong?

Answer 1

You already observed that $\left\Vert x - F_2 \right\Vert = \left\Vert x - F_3 \right\Vert$ which defines a line, the bisector between $F_2$ and $F_3$. This line passes through $\frac{1}{2} F_2 + \frac{1}{2} F_3$ and is orthogonal to $F_3 - F_2$ and is defined by \begin{align*} x_2 &= m \cdot x_1 + b \tag{1} \\ m &= \frac{(F_2)_1 - (F_3)_1}{(F_3)_2 - (F_2)_2} \\ b &=\frac{(F_2)_2 + (F_3)_2}{2} - m \cdot \frac{(F_2)_1 + (F_3)_1}{2} \\ &= \frac{(F_3)_2^2 - (F_2)_2^2 + (F_3)_1^2 - (F_2)_1^2}{2 \left((F_3)_2 - (F_2)_2\right)} \\ &= \frac{\left\Vert F_3 \right\Vert^2 - \left\Vert F_2 \right\Vert^2}{2 \left((F_3)_2 - (F_2)_2\right)} \end{align*}

Now the intersection of the two ellipses has been reduced to the intersection of an ellipse with a line. This is a well known problem. A solution method for this type of problem is described here.

The first ellipse can be rewritten (by squaring) as \begin{align*} \left\Vert x - F_1\right\Vert &= R - \left\Vert x - F_2\right\Vert \\ (x_1 - (F_1)_1)^2 + (x_2 - (F_1)_2)^2 &= R^2 - 2 R \left\Vert x - F_2\right\Vert\\&\phantom{=} + (x_1 - (F_2)_1)^2 + (x_2 - (F_2)_2)^2 \\ 2 ((F_2)_1 - (F_1)_1) x_1 + 2 ((F_2)_2 - (F_1)_2) x_2 + \delta &= -2 R \left\Vert x - F_2\right\Vert \end{align*} where \begin{align*} \delta = (F_1)_1^2 - (F_2)_1^2 + (F_1)_2^2 - (F_2)_2^2 - R^2. \end{align*} Squaring this equation again gives \begin{align} A x_1^2 + B x_1 x_2 + C x_2^2 + D x_1 + E x_2 + F = 0 \tag{2} \end{align} with \begin{align*} A &= 4 ((F_2)_1 - (F_1)_1)^2 - 4R^2 \\ B &= 8 ((F_2)_1 - (F_1)_1) ((F_2)_2 - (F_1)_2) \\ C &= 4 ((F_2)_2 - (F_1)_2)^2 - 4R^2 \\ D &= 4 ((F_2)_1 - (F_1)_1) \delta + 8 R^2 (F_2)_1 \\ E &= 4 ((F_2)_2 - (F_1)_2) \delta + 8 R^2 (F_2)_2 \\ F &= \delta^2 - 4 R^2 ((F_2)_1^2 + (F_2)_2^2). \\ \end{align*} If you insert line equation (1) into equation (2) then you get a quadratic equation which can be easily solved.