# Thread: Find the cartesian form

1. ## Find the cartesian form

Find the cartesian form of r^2=sec(2theta)

I know I have to use conversion formulas x=rcos(theta) and y=rsin(theta) but it's not clear to me how. I've been having trouble with this kind of problem and figure I should get help before I run into more trouble down the line. :s Would appreciate help with this.

2. You have an implicit function in the form $f(r, \theta) = r^{2} - \frac{1}{\cos 2 \theta}=0$ that can be 'transformed' is an implicit function in the form $f(x,y)=0$ setting $r^{2}= x^{2} + y^{2}$ and $\theta = \tan^{-1} \frac{y}{x}$...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by mcsquared
Find the cartesian form of r^2=sec(2theta)

I know I have to use conversion formulas x=rcos(theta) and y=rsin(theta) but it's not clear to me how. I've been having trouble with this kind of problem and figure I should get help before I run into more trouble down the line. :s Would appreciate help with this.
This is the same as

$r^2\cos^2(2\theta)=1 \iff r^2(2\cos^2(\theta)-1)=1 \iff 2(r\cos(\theta))^2-r^2=1$

Now just use what you wrote above

$\displaystyle r^2 = x^2 + y^2$
$\displaystyle \sec{\theta} = \frac{1}{\cos{\theta}}$
$\displaystyle \cos{2\theta} = \cos^2{\theta} - \sin^2{\theta}$
$\displaystyle x = r\cos{\theta} \equiv \cos{\theta} = \frac{x}{r} = \frac{x}{\sqrt{x^2 + y^2}}$
$\displaystyle y = r\sin{\theta} \equiv \sin{\theta} = \frac{y}{r} = \frac{y}{\sqrt{x^2 + y^2}}$.