# How do you graph r=4/(-3+3sintheta)?

##### 1 Answer
Oct 15, 2016

This is a hyperbola with its center $\left(0 , - 4\right)$; its vertices are at $\left(0 , - 4 - \sqrt{12}\right) \mathmr{and} \left(0 , - 4 + \sqrt{12}\right)$ and it opens downward and upward from its vertices.

#### Explanation:

Multiply both sides by the denominator:

$4 r \sin \left(\theta\right) - 3 r = 4$

Multiply both side by -1:

$3 r - 4 r \sin \left(\theta\right) = - 4$

Move the sine term to the right:

$3 r = 4 r \sin \left(\theta\right) - 4$

Factor out a 4:

$3 r = 4 \left(r \sin \left(\theta\right) - 1\right)$

Square both sides:

$9 {r}^{2} = 16 {\left(r \sin \left(\theta\right) - 1\right)}^{2}$

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

$9 \left({x}^{2} + {y}^{2}\right) = 16 {\left(y - 1\right)}^{2}$

Expand the square on the right:

$9 \left({x}^{2} + {y}^{2}\right) = 16 \left({y}^{2} - 2 y + 1\right)$

distribute through the ()s:

$9 {x}^{2} + 9 {y}^{2} = 16 {y}^{2} - 32 y + 16$

Combine like terms and leave the constant on the right:

$9 {x}^{2} - 4 {y}^{2} - 32 y = 16$

Add $- 4 {k}^{2}$ to both sides:

$9 {x}^{2} - 4 {y}^{2} - 32 y - 4 {k}^{2} = 16 - 4 {k}^{2}$

Factor out -4 from the y terms

$9 {x}^{2} - 4 \left({y}^{2} + 8 y + {k}^{2}\right) = 16 - 4 {k}^{2}$

Use 8y to find the value of $k$ and ${k}^{2}$:

$- 2 k y = 8 y$

$k = - 4 , {k}^{2} = 16$

This makes the left a perfect square with k:

$9 {x}^{2} - 4 {\left(y - - 4\right)}^{2} = - 48$

Divide both sides by -48:

${\left(y - - 4\right)}^{2} / 12 - \frac{3}{16} {x}^{2} = 1$

Put in standard form:

${\left(y - - 4\right)}^{2} / {\left(\sqrt{12}\right)}^{2} - {\left(x - 0\right)}^{2} / {\left(4 \frac{\sqrt{3}}{3}\right)}^{2} = 1$

This is a hyperbola with its center $\left(0 , - 4\right)$ its vertices are at $\left(0 , - 4 - \sqrt{12}\right) \mathmr{and} \left(0 , - 4 + \sqrt{12}\right)$. It opens downward and upward from its vertices.