1. ## Changing coordinated

How do I determine the shape polar of the equations as:

$\displaystyle x-3y=0$
$\displaystyle x^2+y^2+4x-2y-5=0$
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.
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2. Originally Posted by Apprentice123
How do I determine the shape polar of the equations as:

$\displaystyle x-3y=0$
$\displaystyle x^2+y^2+4x-2y-5=0$
.
.
.
Substitute $\displaystyle x = r \cos \theta$ and $\displaystyle y = r \sin \theta$ and simplify where required.

3. thanks

4. 1) $\displaystyle y^2+5x=0$

My solution:

$\displaystyle y=psen\theta => y^2=p^2sen^2\theta$
$\displaystyle x=pcos\theta => 5x=5pcos\theta$
$\displaystyle p^2+sen^2\theta+5pcos\theta=0$

$\displaystyle psen^2\theta+5cos\theta=0$

2) $\displaystyle xy=2$

My solution:

$\displaystyle (pcos\theta)(psen\theta)=2$
$\displaystyle p^2cos(\theta).sen(\theta)=2$

$\displaystyle p^2sen2\theta=4$

3) $\displaystyle x^2-4y-4=0$

My solution:

$\displaystyle (pcos\theta)^2-4psen\theta=4$
$\displaystyle p(pcos^2\theta-4sen\theta)=4$

$\displaystyle p=\frac{2}{(1-sen\theta)}$

4) $\displaystyle (x^2+y^2)^2-2a^2xy=0$

My solution:

$\displaystyle p^4=2a^2pcos(\theta).psen(\theta)$

$\displaystyle P^2=a^2sen2\theta$

5. Originally Posted by Apprentice123
1) $\displaystyle y^2+5x=0$

My solution:

$\displaystyle y=psen\theta => y^2=p^2sen^2\theta$
$\displaystyle x=pcos\theta => 5x=5pcos\theta$
$\displaystyle p^2{\color{red}+}sen^2\theta+5pcos\theta=0$

Mr F says: Why the + (in red)?? It should be $\displaystyle {\color{red}\rho^2 \sin^2 \theta + 5 \rho \cos \theta = 0}$.

Then $\displaystyle {\color{red}\rho (\rho \sin^2 \theta + 5 \cos \theta ) = 0}$. Therefore p = 0 or ......

$\displaystyle psen^2\theta+5cos\theta=0$

2) $\displaystyle xy=2$

My solution:

$\displaystyle (pcos\theta)(psen\theta)=2$
$\displaystyle p^2cos(\theta).sen(\theta)=2$

Mr F says: Do you know the double angle formula for $\displaystyle {\color{red}\sin (2 \theta)}$ .....?

$\displaystyle p^2sen2\theta=4$

[snip]

4) $\displaystyle (x^2+y^2)^2-2a^2xy=0$

My solution:

$\displaystyle p^4=2a^2pcos(\theta).psen(\theta)$

Mr F says: See my remarks for 1) and 2) .......

$\displaystyle P^2=a^2sen2\theta$
..

6. Originally Posted by Apprentice123
[snip]
3) $\displaystyle x^2-4y-4=0$

My solution:

$\displaystyle (pcos\theta)^2-4psen\theta=4$
$\displaystyle p(pcos^2\theta-4sen\theta)=4$

$\displaystyle p=\frac{2}{(1-sen\theta)}$
[snip]
$\displaystyle \rho^2 \cos^2 \theta - 4 \rho \sin \theta - 4 = 0$

$\displaystyle \Rightarrow (\cos^2 \theta ) \rho^2 - (4 \sin \theta ) \rho - 4 = 0$

is a quadratic equation in $\displaystyle \rho$. Use the quadratic formula to get (after appropriate simplification) that $\displaystyle \rho = \frac{2}{1 - \sin \theta}$ or $\displaystyle \rho = \frac{-2}{1 + \sin \theta}$.

Note: $\displaystyle \cos^2 \theta = 1 - \sin^2 \theta = (1 - \sin \theta) (1 + \sin \theta)$ will help in the simplifying .....

You should convince yourself that the two solutions above for $\displaystyle \rho$ give the same curve .....

7. thanks