# Finding parameters of an ellipse in terms of Semi-Latus Rectum and Directrix.

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.

• 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? – Ng Chung Tak May 1 '18 at 6:57
• @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 $c$. – Metex May 1 '18 at 19:16
• @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. – 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:

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

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

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

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