Question

Find the polar equation that has the same graph as the equation in x and y:
y^2 - x^2 = 4

Answer 1

Using these to change from x and y to polar:
\[\Large x=r \cos \theta \qquad \qquad y=r \sin \theta\]

\[\Large r^2\sin^2\theta-r^2\cos^2\theta=4\]We can simplify this down a bit,\[\Large r^2\left(\sin^2\theta-\cos^2\theta\right)=4\]Hmm I'm not really sure what we should do with it from here.
I guess we want to write it as r, a function of theta right?
So we could divide both sides by the bracket portion.\[\Large r^2=\frac{4}{\sin^2\theta-\cos^2\theta}\]Square rooting each side,\[\Large r(\theta)=\frac{2}{\sqrt{\sin^2\theta-\cos^2\theta}}\]We could maybe write it terms of just sine or cosine so it looks a little nicer.\[\Large r(\theta)=\frac{2}{\sqrt{2\sin^2\theta-1}}\]

Is this what we're trying to do maybe? :O
Hmm still looks a bit ugly lol

Answer 2

how did you go from \[\frac{ 2 }{ \sqrt{\sin^{2} \theta - \cos^{2} \theta} }\] to \[\frac{ 2 }{ \sqrt{2\sin \theta-1} }\]

Answer 3

and I think its supposed to be in terms of r^2 and sec (theta)

Answer 4

Is it? ah ok let's back up a tad then.

Answer 5

From this step,\[\Large r^2\left(\sin^2\theta-\cos^2\theta\right)=4\]Let's factor a -1 from each term in the brackets,\[\Large -r^2\left(\cos^2\theta-\sin^2\theta\right)=4\]

The `Cosine Double Angle Identity` tells us that:\[\large \color{royalblue}{\cos^2\theta-\sin^2\theta=\cos2\theta}\]

Applying this to our problem:\[\Large -r^2\left(\color{royalblue}{\cos^2\theta-\sin^2\theta}\right)=4\]Gives us,\[\Large -r^2\left(\color{royalblue}{\cos2\theta}\right)=4\]

Answer 6

So I guess from here we would want to isolate the r^2, dividing both sides by -cos2theta,\[\Large r^2=\frac{-4}{\cos2\theta} \qquad\to\qquad r^2=-4\sec2\theta\]

Answer 7

I was trying to avoid doing it this way since we would have to take the root of a negative value :D

But if we want it in terms of r^2, then it's prolly ok.

Answer 8

That works!! Great! THANK YOU!!

Answer 9

using @zepdrix material
\(\bf \color{blue}{sin^2(\theta)-cos^2(\theta) \implies cos(2\theta)}\\
r^2=\cfrac{2}{cos(2\theta)} \implies r^2 = \cfrac{2}{1} \times \cfrac{1}{cos(2\theta)}\)

Answer 10

acck, rather =>
\(\bf \color{blue}{sin^2(\theta)-cos^2(\theta) \implies cos(2\theta)}\\
r^2=\cfrac{4}{cos(2\theta)} \implies r^2 = \cfrac{4}{1} \times \cfrac{1}{cos(2\theta)}\)

Answer 11

hmmm, even that's wrong.... anyhow

Answer 12

$$x=r\cos\theta,y=r\sin\theta\\r^2\sin^2\theta-r^2\cos^2\theta=4\\r^2=-\frac4{\cos^2\theta-\sin^2\theta}=-\frac4{\cos2\theta}$$

Answer 13

Yah I think you have your identity backwards there doe :D\[\large \sin^2\theta-\cos^2\theta \ne \cos2\theta\]

Answer 14

$$\cos^2\theta-\sin^2\theta=\cos2\theta\\\sin^2\theta-\cos^2\theta=-\cos2\theta$$

Answer 15

yes
\(\bf -(cos^2(\theta)-sin^2(\theta)) \implies -cos(2\theta)\\
r^2=\cfrac{2}{-cos(2\theta)} \implies r^2 = \cfrac{2}{1} \times -\cfrac{1}{cos(2\theta)}\)

Answer 16

@jdoe0001 it's \(4\) not \(2\)

Answer 17

yes, my bad, I didn't change that in my editor's version heh
\(\bf -(cos^2(\theta)-sin^2(\theta)) \implies -cos(2\theta)\\
r^2=\cfrac{4}{-cos(2\theta)} \implies r^2 = \cfrac{4}{1} \times -\cfrac{1}{cos(2\theta)}\)

Answer 18

so much for my typos hehe