1
$\begingroup$

A parabola is supposed to sit between ellipse and hyperbola.

And indeed, in the polar form $r=\frac\ell{1+e\cos\theta}$, we pass smoothly through a parabola when the eccentricity $e$ passes through $1$ with a fixed semi latus rectum $\ell$.

But how does it work in cartesian coordinates? Then we have either (in homogeneous coordinates): $$\begin{pmatrix}x&y&1\end{pmatrix} \begin{pmatrix}\frac 1{a^2}\\&\pm\frac 1{b^2}\\&&-1\end{pmatrix} \begin{pmatrix}x\\y\\1\end{pmatrix}=0 $$ Or: $$(a\cos u, b\sin u) \quad\text{versus}\quad (a\cosh u, b\sinh u)$$ While a parabola on the same axis is: $$y^2=4ax \quad\text{respectively}\quad (u^2,\pm\sqrt{4a}u)$$ where $a$ is the semi distance between directrix and focus, similar to the semi long axis of ellipse and hyperbola.

The corresponding homogeneous matrix is: $$\begin{pmatrix}x&y&1\end{pmatrix} \begin{pmatrix}&&-1\\&\frac 1{2a}\\-1\end{pmatrix} \begin{pmatrix}x\\y\\1\end{pmatrix}=0 $$ It doesn't seem to have any similarity to the version of ellipse and hyperbola.

However I look at it, there seems to be a horrible discontinuity/mismatch.

$\endgroup$
  • $\begingroup$ Your parabola should share a vertex of the ellipse and hyperbola, but the vertex of your parabola lies in the center... $\endgroup$ – user376343 Feb 2 at 21:13
  • $\begingroup$ @user376343, we can translate the parabola, but that doesn't seem to help. The homogeneous matrix of the parabola becomes more complicated, and the factor $1/2a$ remains. The parametric $(u^2,\pm\sqrt{4a}u)$ should probably be something like $(1+u^2,\pm\sqrt{4a}u)$ though, so that it is at least in between $\cos$ and $\cosh$ respectively $\sin$ and $\sinh$. $\endgroup$ – Klaas van Aarsen Feb 3 at 13:34
2
$\begingroup$

Here is one possible solution to see the smooth transition (the trick is to choose the right relative sizes of the parameters to get a finite limit object)

Consider a family of ellipses $$ \frac{(x-c)^2}{a^2}+\frac{y^2}{b^2}=1 $$ where $c^2=a^2-b^2$ (hence the origin $x=0$ is at the "left" focus). I will take the limit of the family when $a,b,c\to\infty$ in such a way that $c/a\to 1$ and $b^2/a\to 4p$ (observe that these conditions amount to impose that the limit eccentricity be $1$ and the latus rectum be preserved).

We have $$ y^2=b^2[1-(c/a)^2]+\frac{2cb^2}{a^2}x-(b/a)^2x^2 $$ Given our choice we have $b/a\to 0$, $2cb^2/a^2\to 4p$ and $b^2[1-(c/a)^2]\to 16p^2$. We get in the limit $$ y^2=4px+16p^2=4p(x+4p), $$ a shifted parabola.

$\endgroup$
  • $\begingroup$ You can start with a family of hyperbolas and use a similar argument, with the corresponding modifications $c^2=a^2+b^2$, etc. $\endgroup$ – GReyes Feb 3 at 4:19
1
$\begingroup$

Since you’ve already introduced homogeneous coordinates, this transition can be demonstrated by borrowing some tools from projective geometry.

Consider the one-parameter family of homographies of the plane that fix $(-1,0)$, $(0,1)$ and $(0,-1)$ and map $(1,0)$ to the point with homogeneous coordinates $1:0:\mu$, i.e., to $(1/\mu,0)$ when $\mu\ne0$. In matrix form, this is $$H(\mu) = \begin{bmatrix}1&0&0 \\ 0&\frac{\mu+1}2&0 \\ \frac{\mu-1}2&0&\frac{\mu+1}2\end{bmatrix}.$$ For $\mu\ne-1$, these homographies map the unit circle to the conic $$H^{-T} \operatorname{diag}(1,1,-1) H^{-1} = \begin{bmatrix} {4\mu\over(\mu+1)^2} & 0 & {2(\mu-1)\over(\mu+1)^2} \\ 0 & {4\over(\mu+1)^2} & 0 \\ {2(\mu-1)\over(\mu+1)^2} & 0 & -{4\over(\mu+1)^2}\end{bmatrix},$$ which we can rescale to $$C(\mu) = \begin{bmatrix} \mu & 0 & \frac12(\mu-1) \\ 0&1&0 \\ \frac12(\mu-1)&0&-1\end{bmatrix},$$ i.e., $\mu x^2+y^2+(\mu-1)x=1$.

As $\mu\to0^+$, we get ever-wider ellipses. Informally, we’re stretching the right side of the unit circle farther and farther “toward infinity.” At $\mu=0$, the point $1:0:1$ gets mapped to $1:0:0$—the circle has been “stretched to infinity”—and the resulting conic is the parabola $y^2=x+1$. As $\mu$ becomes negative, the circle can be thought of as continuing to be stretched “beyond” the line at infinity, coming back around the other side, as it were, and the resulting conics are hyperbolas.

This also illustrates the general fact that all nondegenerate conics are projectively equivalent: we’ve generated ellipses, a parabola, and hyperbolas by applying various projective transformations to the unit circle. What determines the type of conic is the number of real intersections with the line at infinity: none, one and two, respectively. However, from a projective-geometric point of view, there’s nothing special about the line at infinity—any line can be chosen for this—so there’s only one type of nondegenerate conic. (For that matter, being able to identify a conic as the unit circle requires some further choices that impose a Euclidean geometry on the projective plane. These choices are implicit in the coordinate system that I’ve used.)

Incidentally, there’s sort of a removable discontinuity at $\mu=-1$. $H(-1)$ is not invertible, so can’t be used directly to construct a conic from the unit circle, but $C(-1)$ is perfectly sensible. To either side of this value the conics are hyperbolas and the vertices of these hyperbolas get closer and closer together as $\mu\to-1$. At $\mu=-1$, the vertices meet: the conic becomes the degenerate hyperbola (pair of lines) $(x+y+1)(x-y+1)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.