How to find the focus of a Cassini oval?

Question

You are given some Cassini oval with foci somewhere on the x-axis. How to construct the foci using a compass and straightedge if only the coordinate axes and the oval are given? The foci are the same distance away from the origin.

This came up while I was doing the following. Given a hyperbola, find all points that see their polar on the hyperbola at a 30° angle. It seems like the set of those points is the Cassini oval.

I conjectured that the line that is both a tangent to the hyperbola and a normal to the Cassini oval intersects one of the axes in the focus of the Cassini oval. Proving this using calculus is very difficult so I was hoping a construction would give some more insight.

The end result should be finding the equation of the Cassini oval for a given hyperbola which means finding the focus first is necessary - hence the question.

Any help is appreciated.

Answer 1

Choose a coordinate system where the foci are $(\pm f,0)$. There are three possibilities.

1. The oval intersect $x$-axis at $4$ points $(\pm u, 0), (\pm v,0 )$ with $u > f > v > 0$. By definition, $(u,0)$ and $(v,0)$ has same product of distances to the foci. This implies $$u^2 - f^2 = |(u-f)(u+f)| = |(f-v)(v+f)| = f^2 - v^2$$

2. The oval intersect $x$-axis at $3$ points $(\pm u, 0), (0,0)$ with $u > f > 0$. We find $$u^2 - f^2 = |(u-f)(f+u)| = |(f-0)(f+0)| = f^2$$

3. The oval intersect $x$-axis at $2$ points $(\pm u,0)$ and $y$-axis at $2$ points $(0,\pm w)$ with $u > f > 0, w > 0$. In this case, $$u^2 - f^2 = \sqrt{(0-f)^2 + w^2}\sqrt{(0+f)^2 + w^2} = f^2 + w^2$$

In these $3$ cases, we have $2f^2 = u^2 + v^2$, $u^2$ and $u^2-w^2$ respectively.

In case 1. We can compute $\sqrt{u^2+v^2}$ by setting up a right triangle with side $u, v$ and look at the hypotenuse.

In case 3. We can compute $\sqrt{u^2-w^2}$ by first setting up a circle of diameter $u$, intersect it with a circle with radius $w$ centered at an endpoint of the diameter and look at the distance between the intersection and the other end point of the diameter.

After these, divide the length $\ell = \sqrt{u^2+v^2}$, $u$ or $\sqrt{u^2-w^2}$ by $\sqrt{2}$ to get $f$. This can be done by setting a square of side $\ell$ and then look at the distance between any vertex and center of the square.

Answer 2

Below are constructions for Cassini ovals with center-to-focus length $a$ and product-of-distances-to-foci $b^2$. In each case, we start with a "circumcircle", locate a closest point $P$ to the center along an axis, then construct points $Q$, $R$, $S$ in that order. From the last of these points we determine an inscribed $2a$-by-$2b$ rectangle, so that we have not only located the foci but also derived the parameter $b$.

• $a > b$:

• $a < b$:

• $a=b$:

Verification is left as an exercise to the reader, who may consider the Cartesian equation $$\left((x-a)^2+y^2\right)\left((x+a)^2+y^2\right)=b^4 $$ and salient algebraic details from @Achille's answer.