Question

Derive the equation (23) at page 42 of Unit 2, which represents the polar equation
of a conic when the directrix L corresponding to a focus F is taken to the right of F.

Answer 1

Let $F$ be a fixed point, the focus, and let $L$ be a fixed line, the directrix, in a plane. A conic section, or conic, is the set of all points $P$ in the plane such that 

$${PF \over PL}=e, e>0$$

where $e$ is a fixed positive number, called the eccentricity.

If $e=1,$ the conic is a parabola.

If $e<1,$ the conic is an ellipse.

If $e>1,$ the conic is a hyperbola.

By locating a focus at the pole, all conics can be represented by similar equations in the polar coordinate system. In each of these equations,

$(r, \theta)$ is a point on the graph of the conic.

$e$ is the eccentricity. (Remember that $e>0$).

$p$ is the distance between the focus (located at the pole) and the directrix.

For a conic with a focus at the origin, if the directrix is $x=\pm p,$ where $p$ is a positive real number, and the eccentricity is a positive real number $e,$ the conic has a polar equation

$$r={e\cdot p \over 1\pm\cos{\theta}}$$

For a conic with a focus at the origin, if the directrix is $y=\pm p,$where $p$ is a positive real number, and the eccentricity is a positive real number $e,$ the conic has a polar equation

$$r={e\cdot p \over 1\pm \sin{\theta}}$$

Given that the directrix $L$ corresponding to a focus $F$ is taken to the right of $F.$

Directrix $L: x=p$

The distance from the focus to the point $P$ in polar is just $r.$

$$PF=r$$

The distance from the point $P$ to the directrix $x=p$ is $PL=p-r\cos{\theta}.$ Then

$$e={PF \over PL}={r \over p-r\cos{\theta}}$$

Solve for $r$

$$e\cdot p-e\cos{\theta}\cdot r=r$$ $$r={e\cdot p \over 1+ \cos{\theta}}$$

The conic has a polar equation

$$r={e\cdot p \over 1+ \cos{\theta}}$$