# Question #27455

Feb 19, 2015

The answer is: ${\left(x - \frac{8}{5}\right)}^{2} / \left(\frac{144}{25}\right) + {y}^{2} / \left(\frac{144}{45}\right) = 1$.

Remembering that:

$x = r \cos \theta$
$y = r \sin \theta$

we can say: ${x}^{2} + {y}^{2} = {r}^{2}$.

Our function becomes:

$r = \frac{4}{3 - 2 \cos \theta} \Rightarrow r \left(3 - 2 \cos \theta\right) = 4 \Rightarrow$

$3 r - 2 r \cos \theta = 4 \Rightarrow 3 \sqrt{{x}^{2} + {y}^{2}} - 2 x = 4 \Rightarrow$

$3 \sqrt{{x}^{2} + {y}^{2}} = 2 x + 4 \Rightarrow$ (with $x \ge - 2$)

$9 {x}^{2} + 9 {y}^{2} = 4 {x}^{2} + 16 x + 16 \Rightarrow$

$5 {x}^{2} - 16 x + 9 {y}^{2} = 16 \Rightarrow$

$5 \left({x}^{2} - \frac{16}{5} x\right) + 9 {y}^{2} = 16 \Rightarrow$

$5 \left({x}^{2} - \frac{16}{5} x + \frac{64}{25} - \frac{64}{25}\right) + 9 {y}^{2} = 16 \Rightarrow$

$5 \left({x}^{2} - \frac{16}{5} x + \frac{64}{25}\right) - \frac{64}{5} + 9 {y}^{2} = 16 \Rightarrow$

$5 {\left(x - \frac{8}{5}\right)}^{2} + 9 {y}^{2} = 16 + \frac{64}{5} \Rightarrow$

$5 {\left(x - \frac{8}{5}\right)}^{2} + 9 {y}^{2} = \frac{144}{5} \Rightarrow$

${\left(x - \frac{8}{5}\right)}^{2} / \left(\frac{144}{5} \cdot \frac{1}{5}\right) + {y}^{2} / \left(\frac{144}{5} \cdot \frac{1}{9}\right) = 1 \Rightarrow$

${\left(x - \frac{8}{5}\right)}^{2} / \left(\frac{144}{25}\right) + {y}^{2} / \left(\frac{144}{45}\right) = 1$

That is an ellipse with center in $C \left(\frac{8}{5} , 0\right)$ and semi-axis $a = \frac{12}{5}$ and $b = \frac{12}{\sqrt{45}} = \frac{12}{3 \sqrt{5}} = 4 \frac{\sqrt{5}}{5}$.