I am trying to solve for all the parameters of an ellipse in terms of the Semi-Latus Rectum, $\ell$, and Directrix, $x$. This is for some equation tables I am making so I am looking for the most simplistic expression. In nearly all the cases it comes down to solving a cubic function.

For example in the case of the linear eccentricity, $c$, I know that: $$ \ell=\frac{\sqrt{c}(c-x)}{\sqrt{x}} $$ Which gets rearranged into: $$ c^3-2c^2 x+c x^2-x \ell^2 = 0 $$ Solving for c using the normal general solution to the cubic equation gives me something nasty. Mathematica also doesn't help. Using a trigonometric solution helps however it gives me a piecewise solution: $$ c =\begin{cases} & \dfrac{2x}{3} \left (1 + sin \left (\frac{arcsin\left (1 - \frac{27 \ell^2}{2 x^2}\right )}{3} \right ) \right )\\ & \dfrac{4x}{3} sin^2 \left (\frac{arccos\left (1 - \frac{27 \ell^2}{2 x^2}\right )}{6} \right ) \end{cases} $$ I feel that this can be simplified further through either assumptions on $x$ and $\ell$ or some trig identity I am unaware of.

If anyone has a reference where they solved for the parameters of an ellipse in terms of just the semi-latus rectum and directrix that would be wonderful. Otherwise any insights on how to reduce the answer further or other methods to attack this problem would be appreciated.

  • $\begingroup$ Poor denotation in your question. We use $\epsilon$, $\varepsilon$ or $e$ to denote eccentricity. Directrices for an ellipse are a pair of lines. Did you mean the directrix to centre distance or directrix to focus distance? $\endgroup$ – Ng Chung Tak May 1 '18 at 6:57
  • $\begingroup$ @NgChungTak I agree that we use that $e$ or $\epsilon$ to denote eccentricity. However, I am talking about linear eccentricity, the distance between the ellipse center and either of its two foci. I have only seen it denoted as $c$. $\endgroup$ – Metex May 1 '18 at 19:16
  • $\begingroup$ @NgChungTak As for your question about the directrix, I mean the distance from center to directrix. Is there another name for this distance? As for the focus to directrix I thought was called the focal parameter of the ellipse. $\endgroup$ – Metex May 1 '18 at 19:19

There's one bug in the first equation though it's unaffected after squaring both sides.

\begin{align} x &= \frac{a^2}{c} \\ a &= \sqrt{cx} \\ \ell &= \frac{b^2}{a} \\ &= \frac{a^2-c^2}{a} \\ &= \frac{cx-c^2}{\sqrt{cx}} \\ \ell \sqrt{x} &= \sqrt{c} \, \color{red}{(x-c)} \\ 0 &= c\sqrt{c}-x\sqrt{c}+\ell \sqrt{x} \\ \sqrt{c} &= 2\sqrt{\frac{x}{3}} \sin \left( \frac{1}{3} \sin^{-1} \frac{3\sqrt{3}\, \ell}{2x}+\frac{2k\pi}{3} \right) \\ c &= \frac{4x}{3} \sin^2 \left( \frac{1}{3} \sin^{-1} \frac{3\sqrt{3}\, \ell}{2x}+\frac{2k\pi}{3} \right) \end{align}

It is quite standard for a cubic equation. The roots are irreducible and also inconstructible by compasses and ruler.

Further points to be noticed:

  • $k=0,1 \implies\sqrt{c}>0 \implies \text{two ellipses}$

  • $k=2 \quad \implies\sqrt{c}<0 \implies \text{a hyperbola}$

  • $\dfrac{X^2}{\frac{4x^2}{3} \sin^2 \left( \frac{1}{3} \sin^{-1} \frac{3\sqrt{3}\, \ell}{2x}+\frac{2k\pi}{3} \right)}+ \dfrac{Y^2}{\frac{2\ell x}{\sqrt{3}} \sin \left( \frac{1}{3} \sin^{-1} \frac{3\sqrt{3}\, \ell}{2x}+\frac{2k\pi}{3} \right)}=1$

  • $e=\dfrac{2}{\sqrt{3}} \left| \sin \left( \dfrac{1}{3} \sin^{-1} \dfrac{3\sqrt{3}\, \ell}{2x}+ \dfrac{2k\pi}{3} \right) \right|$

enter image description here


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.