Chapter 8.5, Problem 55E

Solution

To transform: the polar equation $r=\frac{16}{5-3\cos \theta }$ into cartesian equation $\frac{{\left(x-3\right)}^2}{25}+\frac{y^2}{16}=1$ .

The polar equation is transformed into cartesian equation $\frac{{\left(x-3\right)}^2}{25}+\frac{y^2}{16}=1$ .

Given information:

The polar equation: $r=\frac{16}{5-3\cos \theta }$

Formula used:

  $x=r\cos \theta$ and $x^2+y^2=r^2$

Calculation:

  $r=\frac{16}{5-3\cos \theta }$

Multiply $5-3\cos \theta$ on both sides of the equation.

  $\begin{array}{l}r\left(5-3\cos \theta \right)=16\\ 5r-3r\cos \theta =16\end{array}$

Substitute $r\cos \theta$ for $x$ .

  $\begin{array}{l}5r-3x=16\\ 5r=3x+16\end{array}$

Square both sides of the equation.

  $\begin{array}{l}{\left(5r\right)}^2={\left(3x+16\right)}^2\\ 25r^2=9x^2+96x+256\end{array}$

Substitute $x^2+y^2$ for $r^2$ .

  $\begin{array}{l}25\left(x^2+y^2\right)=9x^2+96x+256\\ 25x^2+25y^2=9x^2+96x+256\end{array}$

Group $x^2$ terms together.

  $\begin{array}{l}25x^2-9x^2-96x+25y^2=256\\ 16x^2-96x+25y^2=256\\ 16\left(x^2-6x\right)+25y^2=256\end{array}$

Use completing the square method to factorize.

  $\begin{array}{l}16\left(x^2-6x+9-9\right)+25y^2=256\\ 16\left({\left(x-3\right)}^2-9\right)+25y^2=256\\ 16{\left(x-3\right)}^2-144+25y^2=256\\ 16{\left(x-3\right)}^2+25y^2=256+144\\ 16{\left(x-3\right)}^2+25y^2=400\end{array}$

Divide both sides of the equation by $400$ .

  $\begin{array}{l}\frac{16{\left(x-3\right)}^2}{400}+\frac{25y^2}{400}=\frac{400}{400}\\ \frac{{\left(x-3\right)}^2}{25}+\frac{y^2}{16}=1\end{array}$

Hence, the polar equation is transformed into cartesian equation $\frac{{\left(x-3\right)}^2}{25}+\frac{y^2}{16}=1$ .