# How do you graph r = 4 / (2+sintheta)?

Oct 23, 2016

Draw an ellipse
The center:$\left(0 , - \frac{4}{3}\right)$
major axis endpoints $\left(- 4 \frac{\sqrt{3}}{3} , - \frac{4}{3}\right) \mathmr{and} \left(4 \frac{\sqrt{3}}{3} , - \frac{4}{3}\right)$
minor axis endpoints $\left(0 , - \frac{8}{3}\right) \mathmr{and} \left(0 , 0\right)$

#### Explanation:

Multiply both sides by the denominator:

$2 r + r \sin \left(\theta\right) = 4$

Substitute $\sqrt{{x}^{2} + {y}^{2}}$ for r and y for $r \sin \left(\theta\right)$:

$2 \sqrt{{x}^{2} + {y}^{2}} + y = 4$

Subtract y from both sides:

$2 \sqrt{{x}^{2} + {y}^{2}} = 4 - y$

Square both sides:

$4 {x}^{2} + 4 {y}^{2} = {\left(4 - y\right)}^{2}$

Expand the square on the right:

$4 {x}^{2} + 4 {y}^{2} = 16 - 8 y + {y}^{2}$

Add $8 y - {y}^{2}$ to both sides:

$4 {x}^{2} + 3 {y}^{2} + 8 y = 16$

Add $3 {k}^{2}$ to both sides:

$4 {x}^{2} + 3 {y}^{2} + 8 y + 3 {k}^{2} = 16 + 3 {k}^{2}$

Change the grouping on the left:

$4 \left({x}^{2}\right) + 3 \left({y}^{2} + \frac{8}{3} y + {k}^{2}\right) = 16 + 3 {k}^{2}$

Find the value of k, and ${k}^{2}$ that completes the square in form:

${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$:

${y}^{2} - 2 k y + {k}^{2} = {y}^{2} + \frac{8}{3} y + {k}^{2}$

$- 2 k y = \frac{8}{3} y$

$k = - \frac{4}{3}$ and ${k}^{2} = \frac{16}{9}$

Substitute ${\left(x - 0\right)}^{2}$ for ${x}^{2}$ and ${\left(y - - \frac{4}{3}\right)}^{2}$ for ${y}^{2} + \frac{8}{3} y + {k}^{2}$ on the left, $\frac{16}{9}$ for ${k}^{2}$ on the right:

$4 {\left(x - 0\right)}^{2} + 3 {\left(y - - \frac{4}{3}\right)}^{2} = 16 + 3 \left(\frac{16}{9}\right)$

Perform the addition on the right:

$4 {\left(x - 0\right)}^{2} + 3 {\left(y - - \frac{4}{3}\right)}^{2} = \frac{64}{3}$

Multiply both sides by $\frac{3}{64}$

$\frac{3}{16} {\left(x - 0\right)}^{2} + \frac{9}{64} {\left(y - - \frac{4}{3}\right)}^{2} = 1$

Write in the standard form of an ellipse:

${\left(x - 0\right)}^{2} / {\left(4 \frac{\sqrt{3}}{3}\right)}^{2} + {\left(y - - \frac{4}{3}\right)}^{2} / {\left(\frac{4}{3}\right)}^{2} = 1$

The center is $\left(0 , - \frac{4}{3}\right)$

Force the y term to zero by setting $y = - \frac{4}{3}$:

${\left(x - 0\right)}^{2} / {\left(4 \frac{\sqrt{3}}{3}\right)}^{2} = 1$

${\left(x - 0\right)}^{2} = {\left(4 \frac{\sqrt{3}}{3}\right)}^{2}$

$x = \pm 4 \frac{\sqrt{3}}{3}$

The endpoints of the major axis are $\left(- 4 \frac{\sqrt{3}}{3} , - \frac{4}{3}\right) \mathmr{and} \left(4 \frac{\sqrt{3}}{3} , - \frac{4}{3}\right)$

For the x term to zero by setting $x = 0$:

${\left(y - - \frac{4}{3}\right)}^{2} / {\left(\frac{4}{3}\right)}^{2} = 1$

${\left(y - - \frac{4}{3}\right)}^{2} = {\left(\frac{4}{3}\right)}^{2}$

$y - - \frac{4}{3} = \pm \frac{4}{3}$

$y = - \frac{4}{3} \pm \frac{4}{3}$

The minor endpoints are $\left(0 , - \frac{8}{3}\right) \mathmr{and} \left(0 , 0\right)$