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Chapter 10: Parametric Equations and Polar Coordinates

Learning module LM 10.6: Conic Sections in Polar Coordinates:

Foci and directrices
Visualizing eccentricity
Polar equations for conic sections
Astronomy

Foci and directrices

Foci and Directrices

Ellipses and hyperbolas are usually defined using two foci, but they can also be defined using a focus and a directrix.
 Definitions of conic sections: Let $L$ be the distance to the focus and let $D$ be the distance to the directrix. Pick a constant $e>0$. The set of all points with $L=eD$ is An ellipse of eccentricity $e$ if $e<1$, A parabola if $e=1$, A hyperbola of eccentricity $e$ if $e>1$.

For instance, suppose that we have a conic section with focus at the origin, directrix at $y=-1$, and eccentricity $e \ne 1$. Then \begin{eqnarray*} \sqrt{x^2 + y^2} &=& e|y+1| \cr x^2 + y^2 & = & e^2 (y^2 + 2 y + 1) \cr x^2 + (1-e^2)y^2 -2e^2 y& = & e^2 . \end{eqnarray*} Notice that the coefficient of $y^2$ in the last equation is positive if $e<1$, giving us an ellipse, and is negative when $e>1$, giving us a hyperbola. After completing the square and applying some more algebraic manipulations, we can put the equation in standard form: $$\left ( \frac{1-e^2}{e^2}\right) x^2 + \left( \frac{(1-e^2)^2}{e^2}\right )\left ( y - \frac{e^2}{1-e^2}\right )^2 = 1.$$

If $e<1$, this is an vertically aligned ("tall and skinny") ellipse with center at $(0,\frac{e^2}{1-e^2})$, with $a = \frac{e}{1-e^2}$, $b=\frac{e}{\sqrt{1-e^2}}$ and $c=\frac{e^2}{1-e^2}$ and eccentricity $e=c/a$. As $e \to 1$, the center and the size of the ellipse both go to infinity.

If $e>1$ the situation is analogous. Our curve is then a hyperbola of with center at $(0,-\frac{e^2}{e^2-1})$, with $a = \frac{e}{e^2-1}$, $b=\frac{e}{\sqrt{e^2-1}}$ and $c=\frac{e^2}{e^2-1}$ and eccentricity $e=c/a$. As $e \to 1$, the center and the size of the hyperbola both go to infinity.