Polar Equations

Hello, Gracy!

You're expected to know the general form for polar conics.

. . $r \:=\:\frac{ke}{1 + e\sin\theta}\qquad\text{or}\qquad r\:=\:\frac{ke}{1 + e\cos\theta}$ . where $e$ = eccentricity

And these rules: . $\begin{array}{ccc} e < 1 & \text{ellipse} \\ e = 1 & \text{parabola} \\ e > 1 & \text{hyperbola}\end{array}$

Quote:

Given the polar equation: . $r\:=\:\frac{4}{2+2\sin x}$
Find the directrix, eccentricity and sketch the conic.

We have: . $r \:=\:\frac{2}{1 + \sin x}\quad\Rightarrow\quad\boxed{e = 1}$

Since $e = 1$ , we have a parabola with its focus is at the pole (origin).

Plot a few points: . $\begin{array}{ccc}x = 0 & r = 2 \\ x = \frac{\pi}{2} & r = 1 \\ x = \pi & r = 2\end{array}$

The graph looks like this:

Code:

2| - - - - - - + - - - - - - | 1| *** * | * * | * ------*---------+---------*------ | * | * |

The focus is 1 unit from the vertex.
. . Hence, the directrix is also 1 unit from the vertex.
It is the horizontal line: $y \:=\:2$
. . Converting to polars: . $r\sin x = 2\quad\Rightarrow\quad\boxed{ r = 2\csc x}$