Q. 64 - Calculus

Question

Let $a\ne 0$. Prove that the graph of the equation $r = \frac{a}{1 - \cos \theta }$ is a parabola in a polar coordinate system. When $a\ne 0$, what is the graph of the equation $r=\frac{a}{1-\sin \theta }$ ? 

Step-by-Step Solution

Answer

It has been proved that graph of equation is a parabola.

The graph of $r=\frac{a}{1-\sin \theta }$ is also a parabola.

Step 1. Given information

$r=\frac{a}{1-\cos \theta }$, $a\ne 0$

Step 2. Convert polar form of equation into rectangular form.

$$\begin{gathered} r=\frac{a}{1-\cos \theta } \\ r=\frac{a}{1-{\displaystyle \frac{x}{r}}} \\ r=\frac{a}{{\displaystyle \frac{r-x}{r}}} \\ r=\frac{ar}{r-x} \\ r-x=a \\ r=x+a \\ \sqrt{x^2+y^2}=x+a \\ {\left(\sqrt{x^2+y^2}\right)}^2={\left(x+a\right)}^2 \\ {x}^2+y^2=x^2+a^2+2ax \\ {y}^2=2ax+a^2 \\ {y}^2=2a\left(x+\frac{a}{2}\right) \end{gathered}$$

This is the equation of parabola which is symmetrical about $x-$axis.

Similarly graph of $r=\frac{a}{1-\sin \theta }$ is also a parabola, which is symmetrical about $y-$axis.