*2016 Dec 27*

After creating a tilted parabola, I started looking into other types of graphs that are also loci of points. A *locus* is a set of all points fulfilling some specific condition, like the parabola, the locus of all points equidistant from the focus point and the directrix line. One simple locus is a circle, which is the locus of all points some given distance from another point, the center of the circle. Another example is the *ellipse*, which is the locus of all points the sum of whose distances from two foci is equal to a constant. Here is an example of what an ellipse might look like:

Another type of locus is the *hyperbola*, which is the locus of all points the difference of whose distances from each of two foci is equal to a constant. Here is a hyperbola:

After studying the ellipse and hyperbola for a only a short time, I began to get bored, because almost everything that there was to know about the ellipse and hyperbola was already discovered. So I decided to make my own locus, which was the locus of all points the product of whose distances from each of two foci was equal to a constant. I decided to let the foci be the points \((x_0,0)\) and \((-x_0,0)\) (so that it would be symmetric about the y-axis) and let this be equation of the locus:

\[\sqrt{y^2+(x-x_0)^2}*\sqrt{y^2+(x-x_0)^2}=c\]

As I played around with this graph, I found that it makes three basic types of shapes. The first consists of two disjoint closed curves that are vaguely egg-shaped,

the second consists of two pointed closed curves sharing a single point,

and the third consists of a single closed curve that is pinched in the middle.

In order to determine when it takes each of these shaped, I decided to define the line equidistant from each of the lemniscate’s foci as the *axis* of the lemniscate. I could then classify each shape by the number of times it intersected its axis: the first shape does not intersect its axis, the second intersects it at \(1\) point, and the third intersects it at \(2\) points. I could then determine algebraically when each shape would occur by setting \(x\) to \(0\) (because when the lemniscate is symmetric about the y-axis, the points at which it intersects its axis are at \(x=0\)). Then I could obtain a simpler expression by solving for the y-position at which the lemniscate would intersect its axis:

\[\sqrt{y^2+(-x_0)^2}*\sqrt{y^2+{x_0}^2}=c\]

\[\sqrt{y^2+{x_0}^2}*\sqrt{y^2+{x_0}^2}=c\]

\[y^2+{x_0}^2=c\]

\[y^2=c-{x_0}^2\]

Now, in order to generalize my findings for any lemnicate not necessarily lying on the x-axis or symmetric about the y-axis, I decided to let \(d\) be the distance between the foci so that \(d=2x_0\) and \({x_0}^2=\frac{1}{4}d^2\), and I replaced

\[y^2=c-{x_0}^2\]

with

\[y^2=c-\frac{1}{4}d^2\]

Then I could easily determine when each of the shapes would be made because

When \(c<\frac{1}{4}d^2\), \(c-\frac{1}{4}d^2\) would be a negative number. This means that \(y^2\) would have to be equal to a negative number, and there would be no real intersections between the lemniscate and its axis.

When \(c=\frac{1}{4}d^2\), \(c-\frac{1}{4}d^2\) would be zero. This means that \(y^2\) would have to be equal to 0, and there would be exactly one intersection (i.e. at \(y=0\)).

When \(c>\frac{1}{4}d^2\), \(c-\frac{1}{4}d^2\) would be a positive number. This means that \(y^2\) would have to be equal to a positive number, and there would be exactly two intersections with ordinates \(y=+\sqrt{c-\frac{1}{4}d^2}\) and \(y=-\sqrt{c-\frac{1}{4}d^2}\).

That’s all I’m doing with the lemniscate for now, but it is a very interesting subject and I will surely return to it later.