Q. 65 - Calculus

Question

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

Step-by-Step Solution

Answer

It has been proved that graph of equation is an ellipse.

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

Step 1. Given information

Equation:

$r=\frac{a}{1-b\cos \theta }$

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

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

Step 3. Simplify the equation by completing square method to make it an equation of ellipse.

$$\begin{gathered} \left(1-b^2\right)\left(x^2-\frac{2abx}{1-b^2}+\frac{y^2}{1-b^2}\right)=a^2 \\ {x}^2-\frac{2abx}{1-b^2}+\frac{y^2}{1-b^2}=\frac{a^2}{1-b^2} \\ \left(x^2-2\times x\times \frac{ab}{1-b^2}+{\left(\frac{ab}{1-b^2}\right)}^2-{\left(\frac{ab}{1-b^2}\right)}^2\right)+\frac{y^2}{1-b^2}=\frac{a^2}{1-b^2} \\ {\left(x-\frac{ab}{1-b^2}\right)}^2+\frac{y^2}{1-b^2}=\frac{a^2}{1-b^2}+{\left(\frac{ab}{1-b^2}\right)}^2 \\ {\left(x-\frac{ab}{1-b^2}\right)}^2+\frac{y^2}{1-b^2}=\frac{a^2(1-b^2)+a^2{b}^2}{{\left(1-b^2\right)}^2} \\ {\left(x-\frac{ab}{1-b^2}\right)}^2+\frac{y^2}{1-b^2}=\frac{a^2-a^2{b}^2+a^2{b}^2}{{\left(1-b^2\right)}^2} \\ {\left(x-\frac{ab}{1-b^2}\right)}^2+\frac{y^2}{1-b^2}=\frac{a^2}{{\left(1-b^2\right)}^2} \\ \frac{{\left(x-\frac{ab}{1-b^2}\right)}^2}{\frac{a^2}{{\left(1-b^2\right)}^2}}+\frac{\frac{y^2}{1-b^2}}{\frac{a^2}{{\left(1-b^2\right)}^2}}=1 \\ \frac{{\left(x-\frac{ab}{1-b^2}\right)}^2}{\frac{a^2}{{\left(1-b^2\right)}^2}}+\frac{y^2}{\frac{a^2}{\left(1-b^2\right)}}=1 \end{gathered}$$

This is the general form of equation of ellipse so the graph of the given equation is an ellipse.

Similarly if we change the equation $r=\frac{a}{1-b\sin \theta }$ into the rectangular coordinate by substituting $\sin \theta =\frac{y}{r}$ then we get an equation which represents also an ellipse.