# How do you graph r=(12)/(3+4 cos theta)?

Nov 24, 2016

#### Explanation:

Multiply both side by $3 + 4 \cos \left(\theta\right)$:

$3 r + 4 r \cos \left(\theta\right) = 12$

Substitute x for $r \cos \left(\theta\right)$:

$3 r + 4 x = 12$

Subtract 4x from both sides:

$3 r = 12 - 4 x$

Square both sides:

$9 {r}^{2} = 144 - 96 x + 16 {x}^{2}$

Substitute $9 {x}^{2} + 9 {y}^{2}$ for $9 {r}^{2}$:

$9 {x}^{2} + 9 {y}^{2} = 144 - 96 x + 16 {x}^{2}$

Move everything but the constant to the left and subtract $7 {h}^{2}$ from both sides:

$- 7 {x}^{2} + 96 x - 7 {h}^{2} + 9 {\left(y - 0\right)}^{2} = 144 - 7 {h}^{2}$

Complete the square for the x term:

$- 7 \left({x}^{2} - \frac{96}{7} x + {h}^{2}\right) + 9 {\left(y - 0\right)}^{2} = 144 - 7 {h}^{2}$

$- 2 h x = - \frac{96}{7} x$
$h = \frac{48}{7}$

$- 7 {\left(x - \frac{48}{7}\right)}^{2} + 9 {\left(y - 0\right)}^{2} = - \frac{1296}{7}$

${\left(x - \frac{48}{7}\right)}^{2} / {\left(\frac{36}{7}\right)}^{2} - {\left(y - 0\right)}^{2} / {\left(12 \frac{\sqrt{7}}{7}\right)}^{2} = 1$

Center$\left(\frac{48}{7} , 0\right)$

${x}_{1} = \frac{48}{7} - \frac{36}{7} = \frac{12}{7}$

${x}_{2} = \frac{48}{7} + \frac{36}{7} = 12$

Vertices: $\left(\frac{12}{7} , 0\right) \mathmr{and} \left(12 , 0\right)$

$\frac{b}{a} = \frac{12 \frac{\sqrt{7}}{7}}{\frac{36}{7}} = \frac{\sqrt{7}}{3}$

asymptotes:

$y = - \left(\frac{\sqrt{7}}{3}\right) \left(x - \frac{48}{7}\right)$

$y = \left(\frac{\sqrt{7}}{3}\right) \left(x - \frac{48}{7}\right)$

Here is the graph: