Chapter 10.3, Problem 77E

To determine

To prove: The polar equation of an ellipse corresponds to the equation in rectangular coordinates $\frac{{\left(x+ae\right)}^2}{a^2}+\frac{y^2}{b^2}=1$ , where $a=\frac{c}{1-e^2}$ and $b=\frac{c}{\sqrt{1-e^2}}$ .

Answer to Problem 77E

It is proved that the polar equation of an ellipse $r=\frac{c}{1+e\cos \theta }$ corresponds to the equation in rectangular coordinates $\frac{{\left(x+ae\right)}^2}{a^2}+\frac{y^2}{b^2}=1$ , where $a=\frac{c}{1-e^2}$ and $b=\frac{c}{\sqrt{1-e^2}}$ .

Explanation of Solution

Given information: The given polar equation of an ellipse is $r=\frac{c}{1+e\cos \theta }$ .

Calculation:

Simplify the given polar equation of an ellipse.

  $\begin{array}{c}r\cdot \left(1+e\cos \theta \right)=c\\ r+re\cos \theta =c\\ r=c-re\cos \theta \\ r^2={\left(c-re\cos \theta \right)}^2\end{array}$

Substitute $x^2+y^2$ for $r^2$ and $x$ for $r\cos \theta$ in the above equation.

  $\begin{array}{c}{x}^2+y^2={\left(c-ex\right)}^2\\ x^2+y^2=c^2+e^2{x}^2-2cex\\ \left(1-e^2\right)x^2+y^2+2cex=c^2\\ \left(1-e^2\right)\cdot \left(x^2+\frac{2cex}{1-e^2}\right)+y^2=c^2\end{array}$

Apply completing square rule and simplify.

  $\begin{array}{c}\left(1-e^2\right)\cdot \left(x^2+2x\frac{ce}{1-e^2}+{\left(\frac{ce}{1-e^2}\right)}^2-{\left(\frac{ce}{1-e^2}\right)}^2\right)+y^2=c^2\\ \left(1-e^2\right)\cdot {\left(x+\frac{ce}{1-e^2}\right)}^2-\left(1-e^2\right)\cdot \frac{c^2{e}^2}{{\left(1-e^2\right)}^2}+y^2=c^2\\ \left(1-e^2\right)\cdot {\left(x+\frac{ce}{1-e^2}\right)}^2+y^2=c^2+\frac{c^2{e}^2}{\left(1-e^2\right)}\\ \left(1-e^2\right)\cdot {\left(x+\frac{ce}{1-e^2}\right)}^2+y^2=\frac{c^2-c^2{e}^2+c^2{e}^2}{1-e^2}\end{array}$

Further simplify the above equation.

  $\begin{array}{c}\left(1-e^2\right)\cdot {\left(x+\frac{ce}{1-e^2}\right)}^2+y^2=\frac{c^2}{1-e^2}\\ \frac{\left(1-e^2\right)\cdot {\left(x+\frac{ce}{1-e^2}\right)}^2}{\frac{c^2}{1-e^2}}+\frac{y^2}{\frac{c^2}{1-e^2}}=1\\ \frac{{\left(x+\frac{c}{1-e^2}\cdot e\right)}^2}{{\left(\frac{c}{1-e^2}\right)}^2}+\frac{y^2}{{\left(\frac{c}{\sqrt{1-e^2}}\right)}^2}=1\end{array}$

Substitute $a$ for $\frac{c}{1-e^2}$ and $b$ for $\frac{c}{\sqrt{1-e^2}}$ in the above equation.

  $\frac{{\left(x+ae\right)}^2}{a^2}+\frac{y^2}{b^2}=1$

Hence, it is proved that the polar equation of an ellipse $r=\frac{c}{1+e\cos \theta }$ corresponds to the equation in rectangular coordinates $\frac{{\left(x+ae\right)}^2}{a^2}+\frac{y^2}{b^2}=1$ , where $a=\frac{c}{1-e^2}$ and $b=\frac{c}{\sqrt{1-e^2}}$ .