Conversion from Polar to Cartesian (ellipse)

1. Nov 4, 2007

nitroracer

1. The problem statement, all variables and given/known data
Convert the conic section to standard form. r=$$\frac{1}{8-4*sin(\theta}$$

2. Relevant equations
x=rcos(\theta)
y=rsinx(\theta)

3. The attempt at a solution

r=$$\frac{1}{8-4*sin(\theta}$$

$$r^2=\frac{1}{64-64*sin(\theta)+16sin^2(\theta)}$$

r^2= x^2 + y^2

I can see the y=r*sin(\theta) but not the x!

2. Nov 5, 2007

Antineutron

so what does r= according to the definitions and what does sin(theta)= according to the definitions? What can you do with that knowledge?

Last edited: Nov 5, 2007
3. Nov 5, 2007

benjyk

I would multiply both sides by (8-4sin(theta)) and then you can replace rsin(theta) with y. And then you can use your identity r^2 = x^2 + y^2.

4. Nov 5, 2007

nitroracer

I came up with $$r= \sqrt{x^2 + y^2}$$ and $$sin(\theta)=y/r$$

$$\sqrt{x^2 + y^2} = \frac{1}{8-4sin(\theta)}$$

$$\sqrt{x^2 + y^2} = \frac{1}{8-(\frac{4y}{r})}$$

$$x^2 + y^2 = 64 - \frac{64y}{r} + \frac{16y^2}{r^2}$$

pretty sure I chose the wrong order of events there...

But when I tried the other suggestion I got pretty close to an answer:

$$8r - 4rsin(\theta) = 1$$

$$8r - 4y = 1$$

$$8r = 1 + 4y$$

$$r = \frac{1+4y}{8}$$

$$r^2 = \frac{1+8y+16y^2}{64}$$

$$x^2 + y^2 = \frac{1+8y+16y^2}{64}$$

$$64x^2 + 64y^2 = 1+8y+16y^2$$

$$64x^2 + 48y^2 - 8y = 1$$

and this is where I get stuck because 48y^2 - 8y +/- ___

5. Nov 5, 2007

r= r/(8r-4y)

8r-4y=1