Changing coordinated

**Apprentice123** · August 30th 2008, 03:36 PM

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

$x-3y=0$
$x^2+y^2+4x-2y-5=0$
.
.
.

**mr fantastic** · August 30th 2008, 03:40 PM

Originally Posted by Apprentice123

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

$x-3y=0$
$x^2+y^2+4x-2y-5=0$
.
.
.

Substitute $x = r \cos \theta$ and $y = r \sin \theta$ and simplify where required.

**Apprentice123** · August 30th 2008, 04:19 PM

thanks

**Apprentice123** · September 2nd 2008, 02:21 PM

1) $y^2+5x=0$

My solution:

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

But the answer is:
$psen^2\theta+5cos\theta=0$

2) $xy=2$

My solution:

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

The answer is:
$p^2sen2\theta=4$

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

My solution:

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

The answer is:
$p=\frac{2}{(1-sen\theta)}$

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

My solution:

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

The answer is:
$P^2=a^2sen2\theta$

**mr fantastic** · September 2nd 2008, 03:56 PM

Originally Posted by Apprentice123

1) $y^2+5x=0$

My solution:

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

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

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

But the answer is:
$psen^2\theta+5cos\theta=0$

2) $xy=2$

My solution:

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

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

The answer is:
$p^2sen2\theta=4$

[snip]

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

My solution:

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

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

The answer is:

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

..

**mr fantastic** · September 2nd 2008, 04:07 PM

Originally Posted by Apprentice123

[snip]
3) $x^2-4y-4=0$

My solution:

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

The answer is:
$p=\frac{2}{(1-sen\theta)}$
[snip]

$\rho^2 \cos^2 \theta - 4 \rho \sin \theta - 4 = 0$

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

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

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

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

**Apprentice123** · September 2nd 2008, 05:07 PM

thanks

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