# What the is the polar form of y^2 = (x-1)^2/y-3x^2y ?

Jun 24, 2018

${r}^{2} = \frac{1}{\sin} ^ 3 \theta \left(\cos \theta \left(r \cos \theta \left(1 + 3 r \left(- 1 + {\cos}^{2} \theta\right) - 2\right)\right) + \frac{1}{r}\right)$

any of the simplifications are answers

#### Explanation:

using the formula

$x = r \cos \theta$

and

$y = r \sin \theta$

substitute

${\left(r \sin \theta\right)}^{2} = \frac{{\left(r \cos \theta - 1\right)}^{2}}{r \sin \theta} - 3 {\left(r \cos \theta\right)}^{2} r \sin \theta$

expand brackets and simplify

${r}^{2} {\sin}^{2} \theta = \frac{{r}^{2} {\cos}^{2} \theta - 2 r \cos \theta + 1}{r \sin \theta} - 3 {r}^{2} {\cos}^{2} \theta r \sin \theta \cdot \frac{r \sin \theta}{r \sin \theta}$

${r}^{2} {\sin}^{2} \theta = \frac{{r}^{2} {\cos}^{2} \theta - 2 r \cos \theta + 1}{r \sin \theta} - \frac{3 {r}^{2} {\cos}^{2} \theta r {\sin}^{2} \theta}{r \sin \theta}$

${r}^{2} {\sin}^{2} \theta = \frac{{r}^{2} {\cos}^{2} \theta - 2 r \cos \theta + 1 - 3 {r}^{2} {\cos}^{2} \theta r {\sin}^{2} \theta}{r \sin \theta}$

simplify

${r}^{2} {\sin}^{2} \theta = \frac{{r}^{2} {\cos}^{2} \theta - 2 r \cos \theta + 1 - 3 {r}^{3} {\cos}^{2} \theta {\sin}^{2} \theta}{r \sin \theta}$

factorise

${r}^{2} {\sin}^{2} \theta = \frac{r \left(r {\cos}^{2} \theta - 2 \cos \theta - 3 {r}^{2} {\cos}^{2} \theta {\sin}^{2} \theta\right) + 1}{r \sin \theta}$

${r}^{2} {\sin}^{2} \theta = \frac{\cancel{r} \left(r {\cos}^{2} \theta - 2 \cos \theta - 3 {r}^{2} {\cos}^{2} \theta {\sin}^{2} \theta\right)}{\cancel{r} \sin \theta} + \frac{1}{r \sin \theta}$

multiply by $\frac{1}{\sin} ^ 2 \theta$

$\frac{1}{\cancel{{\sin}^{2} \theta}} \cdot {r}^{2} \cancel{{\sin}^{2} \theta} = \frac{1}{\sin} ^ 2 \theta \left(\frac{r {\cos}^{2} \theta - 2 \cos \theta - 3 {r}^{2} {\cos}^{2} \theta {\sin}^{2} \theta}{\sin \theta} + \frac{1}{r \sin \theta}\right)$

expand the brackets

${r}^{2} = \frac{r {\cos}^{2} \theta - 2 \cos \theta - 3 {r}^{2} {\cos}^{2} \theta {\sin}^{2} \theta}{{\sin}^{3} \theta} + \frac{1}{r {\sin}^{3} \theta}$

factorise

${r}^{2} = \frac{1}{\sin} ^ 3 \theta \left(r {\cos}^{2} \theta - 2 \cos \theta - 3 {r}^{2} {\cos}^{2} \theta {\sin}^{2} \theta + \frac{1}{r}\right)$

convert ${\sin}^{2} \theta$ to $1 - {\cos}^{2} \theta$

${r}^{2} = \frac{1}{\sin} ^ 3 \theta \left(r {\cos}^{2} \theta - 2 \cos \theta - 3 {r}^{2} {\cos}^{2} \theta \left(1 - {\cos}^{2} \theta\right) + \frac{1}{r}\right)$

expand the $- 3 {r}^{2} {\cos}^{2} \theta \left(1 - {\cos}^{2} \theta\right)$ brackets

${r}^{2} = \frac{1}{\sin} ^ 3 \theta \left(r {\cos}^{2} \theta - 2 \cos \theta - 3 {r}^{2} {\cos}^{2} \theta + 3 {r}^{2} {\cos}^{4} \theta + \frac{1}{r}\right)$

factorise the terms with $r {\cos}^{2} \theta$ in it out

${r}^{2} = \frac{1}{\sin} ^ 3 \theta \left(r {\cos}^{2} \theta \left(1 - 3 r + 3 r {\cos}^{2} \theta\right) - 2 \cos \theta + \frac{1}{r}\right)$

factorise

${r}^{2} = \frac{1}{\sin} ^ 3 \theta \left(\cos \theta \left(r \cos \theta \left(1 + 3 r \left(- 1 + {\cos}^{2} \theta\right) - 2\right)\right) + \frac{1}{r}\right)$