Intersection of two hyperbolas in polar coordinates.

Question

I want to find the intersections of two hyperbolas in polar coordinates. One of their foci coincides, we use this as the pole. (The right focus of the left hyperbola is the same as the left focus of the right hyperbola, as shown on the picture).

I'm trying to solve the following equation to get their intesections:

$$r = \frac{p}{1\mp e \cos \varphi}$$ $$\frac{p_{0}}{1 - e_{0} \cos \varphi} = \frac{p_{1}}{1 + e_{1} \cos \varphi}$$

Hyperbolas can intersect in at most 4 points, but this equation only has two solutions. Where am I going wrong, what am I misunderstanding? How do I find the 4 possible intersections?

Answer 1

Indeed, there is an apparent paradox as can be seen on the following example (similar to yours) with two hyperbolas (featured with their asymptotes) :

$$r=\frac{p_0}{1 - e_0 \cos \varphi} \ \text{and} \ r=\frac{p_1}{1 - e_1 \cos \varphi}\tag{1}$$

with $$e_0=1.5, e_1=2, \ p_0=1, \ p_1=-0.5\tag{2}$$

They have four points of intersection whereas equation

$$\frac{p_0}{1 - e_0 \cos \varphi} = \frac{p_1}{1 - e_1 \cos \varphi}\tag{3}$$

which is equivalent to $6-11 \cos(\varphi)=0$

has only two solutions $\varphi \in (0,2\pi]$.

Remark : Had we chosen any other set of values for $e_0,e_1,p_0,p_1$, we get an equation $a \cos \varphi+b=0$ (similar to (3)) which has at most two solutions on $(0,2 \pi]$.

So why couldn't we get 4 solutions ? What hasn't been taken into account ?

It is due to the fact that, in polar coordinates, a same point can be represented in two ways : by $(r,\varphi)$ or by $\color{red}{(-r,\varphi+\pi)}.$

Therefore, to equation (3) we must add equation :

$$\frac{p_0}{1 - e_0 \cos \varphi} = \color{red}{(-)}\frac{p_1}{1 - e_1 \cos \color{red}{(\varphi+\pi)}}\tag{4}$$

which, in our case, will provide the 2 other solutions.

Remark : please note that the form of the second function in (1) is different from yours : it has $p_1<0$ and a minus sign in the denominator.

Answer 2

The equations have 4 solutions.

Radius vectors have to be considered separately.

Solving separately for the four polar angles we have symmetrical inverse cosine functions for each combination:

$$r_0=\frac{p_{0}}{1 - e_{0} \cos \varphi} ,~ r_2=\frac{p_{1}}{1 - e_{1} \cos \varphi},$$

$$ \cos \varphi_{0+1}=\frac{1-p_0/r_0}{e_0},~ \cos \varphi_{1+0}=\frac{1-p_1/r_1}{e_1}.$$