Curve Tracing

cnpranav · #1 $Old$ October 5th, 2009, 06:44 AM

The equation of the curve is- r^2cos2(theta)=q^2
Sorry I don't know to write this in Natural form but I just hope that I am clear enough

stapel · #2 $Old$ October 5th, 2009, 07:44 AM

What are you supposed to do with this equation?

cnpranav · #3 $Old$ October 5th, 2009, 08:30 AM

Quote:

Originally Posted by stapel $View Post$

What are you supposed to do with this equation?

Make the cure mate..

HallsofIvy · #4 $Old$ October 5th, 2009, 10:56 AM

That's even more confusing!

And what is "q"? Some constant?

Do you know the identity " $cos(2\theta)= cos^2(\theta)- sin^2(\theta)$ "? You can use that to write your equation as $r^2cos^2(\theta)- r^2sin^2(\theta)= q$ and use the fact that $x= r cos(\theta)$ , $y= r sin(\theta)$ to change to Cartesian coordinates. Looks like a hyperbola to me.

cnpranav · #5 $Old$ October 6th, 2009, 12:55 AM

Quote:

Originally Posted by HallsofIvy $View Post$

That's even more confusing!

And what is "q"? Some constant?

Do you know the identity " $cos(2\theta)= cos^2(\theta)- sin^2(\theta)$ "? You can use that to write your equation as $r^2cos^2(\theta)- r^2sin^2(\theta)= q$ and use the fact that $x= r cos(\theta)$ , $y= r sin(\theta)$ to change to Cartesian coordinates. Looks like a hyperbola to me.

Yes q is constant..!!
and the question is $r^2cos^2(\theta)- r^2sin^2(\theta)= q^2$

So I think it will be a rectangular hyperbola..??

Please someone confirm the answer..!!

cnpranav · #6 $Old$ November 8th, 2009, 06:07 AM

Quote:

Originally Posted by stapel $View Post$

What are you supposed to do with this equation?

You have to make the curve with this. By making the curve I mean draw it by the equation

(I was told to give a direct answer to the question)