14.4: The Hyperbola

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Cartesian Coordinates

The hyperbola has eccentricity \(e>1\). In Cartesian coordinates, it has equation

\begin{equation}\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\end{equation}

and has two branches, both going to infinity approaching asymptotes \(x=\pm(a / b) y\). The curve intersects the x axis at \(x=\pm a, \text { the foci are at } x=\pm a e\) for any point on the curve,

\begin{equation}r_{F_{1}}-r_{F_{2}}=\pm 2 a\end{equation}

the sign being opposite for the two branches.

The semi-latus rectum, as for the earlier conics, is the perpendicular distance from a focus to the curve, and is \(\ell=b^{2} / a=a\left(e^{2}-1\right)\). Each focus has an associated directrix, the distance of a point on the curve from the directrix multiplied by the eccentricity gives its distance from the focus.

Polar Coordinates

The \((r, \theta)\) equation with respect to a focus can be found by substituting \(x=r \cos \theta+a e, y=r \sin \theta\) in the Cartesian equation and solving the quadratic for \(u=1 / r\)

Notice that \(θ\) has a limited range: the equation for the right-hand curve with respect to its own focus \(F_{2}\) has

\begin{equation}\tan \theta_{\text {asymptote }}=\pm b / a, \text { so } \cos \theta_{\text {asymptote }}=\pm 1 / e\end{equation}

The equation for this curve is

\begin{equation}\frac{\ell}{r}=1-e \cos \theta\end{equation}

in the range

\begin{equation}\theta_{\text {asymptote }}<\theta<2 \pi-\theta_{\text {asymptote }}\end{equation}

This equation comes up with various signs! The left hand curve, with respect to the left hand focus, would have a positive sign \(\text { + } e\). With origin at \(F_{1}\) (on the left) the equation of the right-hand curve is \(\frac{\ell}{r}=e \cos \theta-1\) finally with the origin at \(F_{2}\) the left-hand curve is \(\frac{\ell}{r}=-1-e \cos \theta\). These last two describe repulsive inverse square scattering (Rutherford).

Note: A Useful Result for Rutherford Scattering

If we define the hyperbola by

\begin{equation}\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\end{equation}

then the perpendicular distance from a focus to an asymptote is just b.

This equation is the same (including scale) as

\begin{equation}\ell / r=-1-e \cos \theta, \text { with } \ell=b^{2} / a=a\left(e^{2}-1\right)\end{equation}

Proof: The triangle \(C P F_{2}\) is similar to triangle \(C H G, \text { so } P F_{2} / P C=G H / C H=b / a\) and since the square of the hypotenuse \(C F_{2} \text { is } a^{2} e^{2}=a^{2}+b^{2}, \text { the distance } F_{2} P=b\)

I find this a surprising result because in analyzing Rutherford scattering (and other scattering) the impact parameter, the distance of the ingoing particle path from a parallel line through the scattering center, is denoted by \(b\). Surely this can’t be a coincidence? But I can’t find anywhere that this was the original motivation for the notation.