# Find intersections of two ellipses who share one fixed point

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?

• Can you add an example sketch? – ja72 Jul 2 '14 at 8:55

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*}
• Since my approach seems not to work well, could you give me a hint on how to translate my equations so that the ellipse is at $(0,0)$ and aligned with the X-axis? I think actually I only need to transform the line, because the ellipse equation does not rely on foci coordinates, right? (only on $a$ and $b$ length as per Wikipedia) – decoromath Jul 2 '14 at 2:48
• I think the math is not completely right. $(x_0, y_0)$ should be calculated as $\frac{1}{2}(F_1 + F_2)$. I do not understand what line you are trying to express by the line equation involving $\vec{p}$ and $\vec{u}$. – philipph Jul 2 '14 at 11:44
• I also noticed that you used $t$ as a parameter for the ellipse and as a parameter for the line. These two parameters are in fact two different parameters. You would have to equate the ellipse and the line to obtain two equations with two unkowns, one for the x-coordinate, one for the y-coordinate. Then you sould be able solve for both parameters without facing the $x = \cos(x)$ problem. – philipph Jul 2 '14 at 11:47