Intersection of two hyperbolas

Question

Let $H$ be a hyperbola defined by $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$$ Let $H'$ be a rotated and shifted hyperbola defined by $$\frac{(x\cos \theta - y\sin \theta + T_x)^2}{c^2} - \frac{(x\sin \theta + y\cos \theta + T_y)^2}{d^2} = 1,$$ where $\theta, T_x, T_y \in \mathbb{R}$.

It is apparently well-known that 2 hyperbolas intersect in at most 4 points, so if $H \neq H'$, we should have $|H \cap H'| \leq 4$. But is there an easy way to rigorously prove this? I have tried substituting $x = \pm a\sqrt{1 + \frac{y^2}{b^2}}$ from the first equation into the second, and after shifting terms and squaring we should obtain the equation

$$a_4y^4 + a_3y^3 + a_2y^2 + a_1y + a_0 = 0,$$

which should have at most 4 solutions. But there are 2 issues:

1. Since $x = \pm a\sqrt{1 + \frac{y^2}{b^2}}$ this leads to at most 8 solutions, instead of 4.
2. What if $a_4 = a_3 = a_2 = a_1 = a_0 = 0$? Due to the squaring it might not necessarily mean that $H \subseteq H'$. I think I can only show that for each $y$, either $\left(a\sqrt{1 + \frac{y^2}{b^2}}, y\right) \in H'$ or $\left(-a\sqrt{1 + \frac{y^2}{b^2}}, y\right) \in H'$, but not necessarily both.

Answer 1

This can be seen as a special case of Bézout's theorem.

In general you have to show two things, namely that you can indeed have four points of intersection. An example would suffice for this. And secondly that if two conics have more than four points in common, then they are the same. Which is why @mathlove wrote a comment about Five points determine a conic. Actually this part is only true for non-degenerate conics. If you consider a pair of lines as a special case of conic (since you can write a quadratic equation for it, simply as the product of two linear equations), then you can have two pairs of lines which have a whole line in common. So the general statement is that two conics have $4$ points in common unless they have a whole component in common, with component being a factor of the defining polynomial. Which must be the whole polynomial in the non-degenerate i.e. irreducible case.

Now let's look at your considerations.

1. Since $x = \pm a\sqrt{1 + \frac{y^2}{b^2}}$ this leads to at most 8 solutions, instead of 4.

   Yes, you have two $x$ coordinates to consider for every $y$ coordinate, but actually only one of them will lie on the second conic.

2. What if $a_4 = a_3 = a_2 = a_1 = a_0 = 0$? Due to the squaring it might not necessarily mean that $H \subseteq H'$. I think I can only show that for each $y$, either $\left(a\sqrt{1 + \frac{y^2}{b^2}}, y\right) \in H'$ or $\left(-a\sqrt{1 + \frac{y^2}{b^2}}, y\right) \in H'$, but not necessarily both.

   You are partially right. The situation where all four coefficients are zero but the conics are still not equal can occur in the case of degenerate conics, as described above. However, the conics described by your formulas are non-degenerate for $a,b,c,d\neq 0$ and undefined if one of them should be zero.

   I'd again prove this by demonstrating that if two conics have five points in common, no thee of them collinear, then they are the same conic.

Answer 2

Concerning your computation, as you are likely to enlarge the set of solutions by successive squarings, I do not find astonishing that spurious solutions appear.

I can give a simple explanation about the fact that any given hyperbola intersects another hyperbola in not more than four points.

In fact, up to an affine transform, one can write the equation of the hyperbola under the form

$$\tag{1}y=\frac{1}{x}.$$

Let the other conic be written under the general form:

$$\tag{2}ax^2+2bxy+cy^2+2dx+2ey+f=0$$

Plugging (1) into (2) will result in a fourth degree equation in $x$ with a maximum of 4 real solutions. To each real solution $x_0$, one can associate only one $y_0=\frac{1}{x_0}$.