Question

Derive equation (c) in Table 5:
$r=\frac{ep}{1+e\sin \theta }$

Short answer

The polar equation$r=\frac{ep}{1+e\sin \theta }$has been proved.

Step-by-step solution

Step 1. Given information

We have to derive the following equation :

$r=\frac{ep}{1+e\sin \theta }$

Step 2. Examine the given polar equation

To do that, let us examine first the given polar equation. We can observed that the denominator has $a+\sin \theta$term which means that the directrix is parallel to the polar axis and is located p units above to the right of the pole.

Step 3. Find the value of D

Based from the figure,

$D=p-l$

Find the value of l,

$\sin \theta =\frac{l}{r}\Rightarrow l=r\sin \theta$

Substitute the value,

$D=p-r\sin \theta$

Step 4. Derive the equation

Relation between radius and eccentricity is given by :

$r=eD$

Apply previous equation to the equation of r,

$\begin{array}{r}r=e(p-r\sin \theta )\\ r\stackrel{\left(1\right)}{=}ep-er\sin \theta \\ r+er\sin \theta \stackrel{\left(2\right)}{=}ep\\ r(1+e\sin \theta )\stackrel{\left(3\right)}{=}ep\\ r\stackrel{\left(4\right)}{=}\frac{ep}{1+e\sin \theta }\end{array}$