Chapter 8.5, Problem 37E

Solution

To find: The Cartesian equation for the given polar equation.

The Cartesian equation for the given polar equation is $\frac{9{\left(y-\frac{4}{3}\right)}^2}{64}+\frac{3x^2}{16}=1$ .

Given information:

The given equation is $r=\frac{4}{2-\sin \theta }$ .

Formula used:

The formula to calculate the value of $c$ is given by,

  $c=ea$   ..... (1)

The formula to calculate the value of $b$ is given by,

  $\begin{array}{c}{a}^2=b^2+c^2\\ b=\sqrt{a^2-c^2}\end{array}$   ..... (2)

Calculation:

Divide the numerator and denominator by $2$ .

  $r=\frac{1}{1-\frac{1}{2}\sin \theta }$

Compare the above equation with $r=\frac{ke}{1-e\sin \theta }$ .

  $\begin{array}{c}e=\frac{1}{2}\\ <1\left(\text{ellipse}\right)\end{array}$

The focal axis is the line $\theta =\frac{\pi }{2}$ (or y -axis) since $\sin \theta$ is present in the polar equation so find the endpoints of the major axis using $\theta =\frac{\pi }{2}$ and $\theta =\frac{3\pi }{2}$ .

When $\theta =\frac{\pi }{2}$ ,

  $\begin{array}{c}r=\frac{4}{2-\sin \frac{\pi }{2}}\\ =\frac{4}{2-1}\\ =4\end{array}$

When $\theta =\frac{3\pi }{2}$ ,

  $\begin{array}{c}r=\frac{4}{2-\sin \frac{3\pi }{2}}\\ =\frac{4}{2+1}\\ =\frac{4}{3}\end{array}$

The endpoints of the major axis are $\left(4,\frac{\pi }{2}\right)$ and $\left(\frac{4}{3},\frac{3\pi }{2}\right)$ so the major axis length is as follows.

  $\begin{array}{c}2a=4+\frac{4}{3}\\ 2a=\frac{16}{3}\\ a=\frac{8}{3}\end{array}$

Substitute $\frac{1}{2}$ for $e$ and $\frac{8}{3}$ for $a$ in equation (1) to find $c$ .

  $\begin{array}{c}c=\left(\frac{1}{2}\right)\left(\frac{8}{3}\right)\\ =\frac{4}{3}\end{array}$

Substitute $\frac{8}{3}$ for $a$ and $\frac{4}{3}$ for $c$ in equation (2) to find $b$ .

  $\begin{array}{c}b=\sqrt{{\left(\frac{8}{3}\right)}^2-{\left(\frac{4}{3}\right)}^2}\\ =\sqrt{\frac{64}{9}-\frac{16}{9}}\\ =\sqrt{\frac{16}{3}}\\ =\sqrt{6}\end{array}$

In the Cartesian plane, the focal axis is the y -axis and the endpoints of the major axis are $\left(0,4\right)$ and $\left(0,-\frac{4}{3}\right)$ so the center is at:

  $\begin{array}{c}\left(h,k\right)=\left(\frac{0+0}{2},\frac{4-\frac{4}{3}}{2}\right)\\ =\left(0,\frac{4}{3}\right)\end{array}$

Using the standard form of a vertical ellipse, the equation is as follows.

  $\begin{array}{c}\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1\\ \frac{{\left(y-\frac{4}{3}\right)}^2}{{\left(\frac{8}{3}\right)}^2}+\frac{{\left(x-0\right)}^2}{{\left(\frac{\sqrt{16}}{3}\right)}^2}=1\\ \frac{{\left(y-\frac{4}{3}\right)}^2}{\frac{64}{9}}+\frac{x^2}{\frac{16}{3}}=1\\ \frac{9{\left(y-\frac{4}{3}\right)}^2}{64}+\frac{3x^2}{16}=1\end{array}$

Therefore, the Cartesian equation for the given polar equation is $\frac{9{\left(y-\frac{4}{3}\right)}^2}{64}+\frac{3x^2}{16}=1$ .