r=8/1-cos theta
Can someone please help me understand this problem.
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First, your second line is not written well. You just implied that $\displaystyle r = r^2$. Please be careful with your equal signs!
Let's start with line 2. You have (rewriting a little bit):
$\displaystyle r = 8 \frac{1 + cos( \theta )}{1 - cos^2 ( \theta )} = 8 \frac{1 + cos( \theta )}{sin^2( \theta )}$
We want to get that $\displaystyle sin^2( \theta ) $ to go to $\displaystyle r^2 ~ sin^2 ( \theta )$ so we multiply the top and bottom by $\displaystyle r^2$:
$\displaystyle = 8 \frac{r^2 (1 + cos( \theta ))}{r^2 ~ sin^2( \theta )}$
We need the $\displaystyle cos( \theta )$ to go to $\displaystyle r~cos( \theta )$, so let's take one of those r's in the numerator and multiply it through:
$\displaystyle = 8 \frac{r (r + r~cos( \theta ))}{r^2 ~ sin^2( \theta )}$
Now we do the x, y thing:
$\displaystyle = 8 \frac{r (r + x)}{y^2}$
Or going back to the beginning of line 2:
$\displaystyle r = \frac{r (r + x)}{y^2}$
Canceling the r's we get:
$\displaystyle 1 = 8 \frac{(r + x)}{y^2}$
The problem with this even with all this work we still have an r in there.
Let's rethink this. We know that $\displaystyle r = \sqrt{x^2 + y^2}$ and $\displaystyle cos ( \theta ) = \frac{x}{r} = \frac{x}{\sqrt{x^2 + y^2}}$ So the original equation in line one reads:
$\displaystyle \sqrt{x^2 + y^2} = 8 \frac{1}{1 - \frac{x^2}{x^2 + y^2}}$
Can you finish from there?
-Dan