# Curve of a fixed point of a conic compelled to pass through 2 points

Suppose that in the plane a given conic curve is compelled to pass through two fixed points of that plane.

What are the curves covered by a fixed point of the conic, its center (for an ellipse), its focus, etc. ?

(I apologize for the bad English ...)

-
Your question is ill-posed. To restrict to just only ellipses, for example, one can draw infinitely many ellipses in different orientations that pass through two points. – J. M. Nov 22 '10 at 8:50
On the other hand, if you're asking about the locus of a fixed point of a conic of preset type/dimensions constrained to slide on two fixed points, then you have an answerable question... – J. M. Nov 22 '10 at 8:55
@J.M. Good distinction. But can't we assume "a given conic curve" means the same thing as "preset type/dimensions" and that only proper Euclidean motions of this curve are contemplated? – whuber Nov 22 '10 at 16:37
Or, perhaps the intent is to fix the shape (that is, the eccentricity) of the conic, in an investigation of possible generalizations of the observation that "the locus of centers of all circles passing through two points is the perpendicular bisector of the segment joining those points". The locus of foci of co-eccentric conics through two given points contains at least a subset of the perpendicular bisector, via conics with major axes perpendicular to the segment. What pts are added when the conics' axes are allowed to be "skew" to the perpendicular? In particular, what about parabolas? – Blue Nov 22 '10 at 18:31
As OP didn't respond to J. M.'s request for clarification, I vote to close as not a real question – Ross Millikan Apr 29 '11 at 16:55

First, a few animations:

These were generated by a parabola with focal length $a=1$ and distance between two points $c=5$. The first one has the focus of the parabola as the tracing point, while the second one has the vertex as tracing point.

Now, for the mathematics: using whuber's and Don's comments as a possible interpretation, the question is asking what would be the (point-)glissette of a conic sliding between two points. I'll consider the parabolic case here, since it's the easiest of the three.

Here's the general idea: start with some parabola $(2at\quad at^2)^T$ (where $a$ is the "focal length", or the distance from the vertex to the focus), imagine a chord of length $c$ moving along the parabola, and then translate and rotate the parabola in such a way as to have the chord's endpoints match the two fixed points at $(-c/2,0)$ and $(c/2,0)$.

Here's the complication: letting the two points on the parabola at a distance $c$ from each other have the parameters $u$ and $u+h$, we obtain the quartic equation

$$h^4+4uh^3+4(1+u^2)h^2-\left(\frac{c}{a}\right)^2=0$$

and as you might know, solving a quartic equation is complicated. The algebra is hellish, and I'll thus skip that for the time being. Assuming that we now have the (complicated!) function $h(a,c,u)$ for computing the lone positive root of that quartic equation, here's what you do: translate the tracing point $(x_t,y_t)$ so that the point $(2au\quad au^2)^T$ is the origin, rotate by an appropriate rotation matrix, and then translate again by the point $(c/2,0)$. The "appropriate matrix" is obtained by considering the slope of the chord of length $c$ of the parabola:

$$m=\frac{a(u+h)^2-au^2}{2a(u+h)-2au}=u+\frac{h(a,c,u)}{2}$$

and from that construct the rotation matrix

$$\frac1{\sqrt{1+m^2}}\begin{pmatrix}1&m\\-m&1\end{pmatrix}$$

Assembling that all together gives

$$\frac1{\sqrt{1+(u+h(a,c,u)/2)^2}}\begin{pmatrix}1&u+h(a,c,u)/2\\-u-h(a,c,u)/2&1\end{pmatrix}\cdot\left(\begin{pmatrix}x_t\\y_t\end{pmatrix}-\begin{pmatrix}2au\\au^2\end{pmatrix}\right)-\begin{pmatrix}c/2\\0\end{pmatrix}$$

You can obtain the complicated parametric equations for the parabola glissette by replacing the $h(a,c,u)$ with the appropriate expression for the positive root of the quartic equation given earlier.

The elliptic and hyperbolic cases are even more complicated than this; I'll leave the investigation of that to someone with more endurance and mathematical ability than me. :)

Here is a Mathematica notebook for generating these animations.

In the Mathematica notebook I provided, I used the function Root[] for representing the function $h(a,c,u)$. To show that I wasn't pulling the leg of you, the gentle reader, I'll display the explicit form of $h(a,c,u)$, the way Ferrari would've.

Consider again the quartic equation

$$h^4+4uh^3+4(1+u^2)h^2-\left(\frac{c}{a}\right)^2=0$$

The resolvent cubic for this quartic is

$$y^3-4(u^2+1)y^2+\frac{4c^2}{a^2}y-\frac{16c^2}{a^2}=0$$

and the (only) positive root of this cubic is given by the expression

$$y_+=\frac13\left(4(1+u^2)+\frac{2(4a^2 (1+u^2)^2-3c^2)}{a\sqrt[3]{v}}+\frac{2}{a}\sqrt[3]{v}\right)$$

where

$$v=8a^3 (1+u^2)^3-9ac^2 (u^2-2)-3c\sqrt{3}\sqrt{c^4+16a^4 (1+u^2)^3+a^2 c^2 (8-20u^2-u^4)}$$

and the real cube root is always taken.

From $y_+$, we can compute $h(a,c,u)$ as the positive root of the quadratic

$$h^2+\frac{h}{2}(4u-2\sqrt{y_+-4})+\frac12\left(y_+-\sqrt{\frac{4c^2}{a^2}+y_+^2}\right)$$

that is,

$$h(a,c,u)=-u+\frac{\sqrt{y_+-4}}{2}+\sqrt{u^2-1-u\sqrt{y_+-4}-\frac{y_+}{4}+\sqrt{\frac{y_+^2}{4}+\frac{c^2}{a^2}}}$$

(I told you it was complicated... ;))

-