Conics

Section 8.1 Conics

Subsection 8.1.1

We are going to need conics to talk about orbital mechanics in the next section. Let's first review the classical definitions of the four types of conics.

Definition 8.1.1.

The circle is all points equidistant from a center point (focus). It is determined entirely by its centre point and radius.

\begin{align*} \amp \text{ Parameters } \amp \amp \text{ Radius } r \\ \amp \text{ Equation at } (0,0) \amp \amp x^2 + y^2 = r^2 \\ \amp \text{ Equation at } (x_0, y_0) \amp \amp (x-x_0)^2 + (y-y_0)^2 = r^2 \end{align*}

Definition 8.1.2.

The ellipse is all points equidistant from two points (foci). For the ellipse centered at the origin, the foci are \((\pm c,0)\) where \(c^2 = a^2 - b^2\text{.}\) The ellipse is determined by a centre point and two axis length, or by the two foci and a distance. (This description of the ellipse assumes \(a > b\text{.}\) If \(a \lt b\text{,}\) the semimajor and semiminor axes are switched, since the major axis is always larger, the foci would be on the \(y\) axis instead of the \(x\text{,}\) axis and \(c^2 = b^2 - a^2\text{.}\))

\begin{align*} \amp \text{ Parameters } \amp \amp \text{ Axes } a, \ \ b \\ \amp \text{ Equation at } (0,0) \amp \amp \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \\ \amp \text{ Equation at } (x_0, y_0) \amp \amp \frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1 \end{align*}

Definition 8.1.3.

The parabola is all points which are equidistant to a point (focus) and a line (directrix). For the parabola centered at the origin, the focus is \((0,p)\) and the directrix is the line \(y = -p\text{.}\) The focus and directrix entirely determine the parabola. (This description holds for upward opening parabolae. If the directrix is \(y = p\) and the focus \((0, -p)\text{,}\) then \(x^2 = -4py\) is the downward facing parabola. Switching the roles of \(x\) and \(y\) gives left or right facing parabolae.)

\begin{align*} \amp \text{ Parameters } \amp \amp \text{ Focus - Directrix Distance } 2p \\ \amp \text{ Equation at } (0,0) \amp \amp x^2 = 4py\\ \amp \text{ Equation at } (x_0, y_0) \amp \amp (x-x_0)^2 = 4p(y-y_0) \end{align*}

Definition 8.1.4.

The hyperbola is all points where the difference of the distances to the two foci is constant. The foci, similar to the ellipse, are the points \((\pm c,0)\) with \(c^2 = a^2 + b^2\text{.}\) The hyperbola has vertices \((\pm a,0)\text{,}\) which are the points closest to the origin. In addition, it has asymptotes, which are the lines \(y = \pm \frac{b}{a} x\text{.}\) The hyperbola is determined by its two foci and a distance. (This description is for hyperbolae which open along the \(x\) axis. Switching \(x\) and \(y\) gives hyperbolae which open along the \(y\) axis.)

\begin{align*} \amp \text{ Parameters } \amp \amp \text{ Axes } a, \ b\\ \amp \text{ Equation at } (0,0) \amp \amp \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \\ \amp \text{ Equation at } (x_0, y_0) \amp \amp \frac{(x-x_0)^2}{a^2} - \frac{(y-y_0)^2}{b^2} = 1 \end{align*}

Subsection 8.1.2 Eccentricity

The parabola is defined as all points equidistant from a focus and a directrix. All other conics have foci, but the parabola is the only with a directrix. However, there is another method of describing all conics in terms of focus and directrix. To get to that description, consider first the parabola in more detail in the following diagram. In Figure 8.1.5, we've labeled the conic itself as \(\gamma\text{,}\) the focus \(F\) and the directrix \(L\) and the distance between the focus and the directrix, labeled \(d\text{.}\)

Figure 8.1.5. Conic definition with directrix \(L\) and focus \(F\)

We've also labeled an angle \(\theta\) and two distances \(r\) and \(l\) in this diagram. The point \(P\) is, by definition, on the parabola because the distances from \(P\) to the focus and to the directrix are equal. In Figure 8.1.5, \(r=l\text{.}\)

We can generalize the construction by allowing \(r\) and \(l\) to be non-equal, but insisting that their ration is constant. That is, given a focus \(F\) and a directrix \(L\text{,}\) we consider all points \(P\text{,}\) such \(\frac{r}{l}\) is a constant. \(\frac{r}{l} = 1\) will recover the parabola, but other choices of the constant will result, amazingly, in other conics.

Definition 8.1.6.

The eccentricity of a conic defined via a focus and a directix is the ratio \(\frac{r}{l}\) of the distance to the focus over the distance to the directrix.

\begin{equation*} e = \frac{r}{l} \end{equation*}

We would like to get a description of conics as a parametric curves using the focus, directrix and eccentricity. We will use polar coordinates for this, assuming that \(F\) is at the origin. In Figure 8.1.5, putting \(F\) at the origin means that \(r\) and \(\theta\text{,}\) as labelled, are exactly polar coordinates.

Proposition 8.1.7.

Let \(\gamma\) be a conic described by a focus at the origin, a directrix \(x=d\) and an eccentricity \(e\text{.}\) Then \(\gamma\) is described by a polar locus,

\begin{equation*} r = \frac{ed}{1 + e \cos \theta}\text{,} \end{equation*}

or by a polar parametric curve using \(\theta\) as the parameter,

\begin{equation*} \gamma(\theta) = (r(\theta), \theta) = \left(\frac{ed}{1 + e \cos \theta}, \theta\right) \end{equation*}

Proof.

From the diagram and some trigonometry, we can see that \(l = d - r\cos \theta\text{.}\) Eccentricity is the ratio of \(r\) to \(l\text{,}\) so eccentricity has this form:

\begin{equation*} e = \frac{r}{d-r\cos\theta} \end{equation*}

If we solve for \(r\) in this equaiton we get the desired polar locus.

\begin{equation} r = \frac{ed}{1 + e \cos \theta}\label{equation-conic-eccentricity-form}\tag{8.1.1} \end{equation}

Then we simply choose \(\theta\) as the parameter for the polar parametric description.

Proposition 8.1.8.

All four conics can be defined in terms of a focus \(F\text{,}\) directrix \(L\text{,}\) and eccentricity \(e\text{.}\) A conic is thus redefined to be all points where the ratio of the distance to the focus over the distance to the directrix is constant and equal to the eccentricity. The hyperbola has eccentricity \(e>1\text{,}\) the parabola has eccentricity \(e=1\text{,}\) the ellipse has eccentricity \(e\lt 1\text{.}\) The circle is the special limit case \(e=0\text{,}\) though to realize the shape we have to allow the distance between \(F\) and \(L\) to approach infinity.

Proof.

We will prove all cases except the limit case of the circle. Our approach is to start with the polar locus and try to recover the standard equations of the conics from the start of this section.

We start with the polar locus and change back to Cartesian coordinates, using the change of coordinates equations: \(r = \sqrt{x^2 + y^2}\) and \(r \cos \theta = x\text{.}\)

\begin{align*} \frac{r}{d - r\cos \theta} \amp = e\\ \frac{\sqrt{x^2 + y^2}}{d - x} \amp = e\\ \sqrt{x^2 + y^2} \amp = ed -ex = e(d-x)\\ x^2 + y^2 \amp = e^2 (d^2 - 2dx + x^2) \end{align*}

We rearrange this to get a polynomial expression in \(x\text{.}\)

\begin{equation} (1-e^2)x^2 + 2de^2 x + y^2 = d^2 e^2\label{equation-reference3}\tag{8.1.2} \end{equation}

If \(e \neq 1\text{,}\) we divide through by \((1-e^2)\text{,}\) then proceed to complete the square in the \(x\) term.

\begin{align*} x^2 + \frac{2de^2}{1-e^2} x + \frac{y^2}{1-e^2} \amp = \frac{d^2 e^2}{1-e^2}\\ x^2 + \frac{2de^2}{1-e^2} x + \frac{d^2e^4}{(1-e^2)^2} - \frac{d^2e^4}{(1-e^2)^2} + \frac{y^2}{1-e^2} \amp = \frac{d^2 e^2}{1-e^2}\\ \left( x + \frac{e^2d}{1-e^2} \right)^2 + \frac{y^2}{1-e^2} \amp = \frac{d^2 e^2}{1-e^2} + \frac{d^2e^4}{(1-e^2)^2}\\ \end{align*}

We take the right-hand side to common denominator.

\begin{align*} \left( x + \frac{e^2d}{1-e^2} \right)^2 + \frac{y^2}{1-e^2} \amp = \frac{d^2 e^2(1-e^2) + d^2e^4}{(1-e^2)^2}\\ \left( x + \frac{e^2d}{1-e^2} \right)^2 + \frac{y^2}{1-e^2} \amp = \frac{d^2 e^2}{(1-e^2)^2}\\ \end{align*}

We divide by the right-hand side.

\begin{align*} \frac{\left( x + \frac{e^2d}{1-e^2} \right)^2}{\frac{d^2e^2}{(1-e^2)^2}} + \frac{\frac{y^2}{1-e^2}}{\frac{d^2e^2}{(1-e^2)^2}} \amp = 1\\ \frac{\left( x + \frac{e^2d}{1-e^2} \right)^2}{\frac{d^2e^2}{(1-e^2)^2}} + \frac{y^2}{\frac{d^2e^2}{1-e^2}} \amp = 1\\ \end{align*}

The sign in from of the \(y^2\) term is positive if \(1-e^2 > 1\text{,}\) that is, when \(e\lt 1\text{.}\) Otherwise, \(e^2-1\) is positive. In the positive case, we can rewrite the expression as a difference of squares.

\begin{align*} \frac{\left( x + \frac{e^2d}{1-e^2} \right)^2}{\frac{d^2e^2}{(1-e^2)^2}} - \frac{y^2}{\frac{d^2e^2}{e^2-1}} \amp = 1 \end{align*}

In the case \(e \lt 1\text{,}\) the expression is the sum of positive squares, so we have an ellipse. In the case \(e>1\text{,}\) the expression is the difference of positive square, so we have a hyperbola. For future convenience, we define the following constants for the \(e\lt 1\) (ellipse) case.

\begin{equation} x_0 = \frac{-e^2 d}{1-e^2} \label{equation-reference7-1}\tag{8.1.3} \end{equation}

\begin{equation} a^2 = \frac{e^2 d^2}{(1-e^2)^2}\label{equation-reference7-2}\tag{8.1.4} \end{equation}

\begin{equation} b^2 = \frac{e^2 d^2}{1-e^2}\label{equation-reference7-3}\tag{8.1.5} \end{equation}

Subsituting these cosntant into the previous equation gives an elegant ellipse equation.

\begin{equation*} \frac{(x-x_0)^2}{a^2} \pm \frac{y^2}{b^2} = 1 \end{equation*}

Lastly, in Equation (8.1.2) we excluded \(e = 1\text{.}\) Now assume \(e=1\text{,}\) which removes the \((1-e^2)x^2\) term from Equation (8.1.2) entirely. We continue the derivation from that point; the calculation is much easier in this case.

\begin{equation*} 2dx + y^2 = d^2 \implies y^2 = d^2 - 2dx \implies y^2 = -2d \left( x - \frac{d}{2} \right) \end{equation*}

This is the equation for a leftward-opening parabola, so \(e=1\) does recover the parabola.

For all cases except the circle, we recovered the standard equation of the conic. We found a hyperbola when \(e>1\text{,}\) an ellipse when \(e\lt 1\text{,}\) and a parabola when \(e=1\text{.}\) As mentioned in the proposition, the circle is a strange limit case as \(e \rightarrow 0\text{,}\) but \(e \rightarrow 0\) also increases the distance \(d \rightarrow \infty\text{.}\)