The Student Room Group
ssmoose
r^2 = sec(2x)

and

r^2 = cosec(2x)


r^2=cosec(2x)
r^2sin2x=1
2r^2sinxcosx=1
2(rsinx)(rcosx)=1


r^2=sec(2x)
r^2cos(2x)=1
r^2(cos^2x-sin^2x)=1
(rcosx)^2-(rsinx)^2=1

edit: x is a bad choice for theta !
ssmoose
1.) r^2 = sec(2x)
2.) r^2 = cosec(2x)

1.) x = rcost, y = rsint.
r^2 = sec(2x) = 1/(cos2x) = x^2/cos^2t

sint = y/r => 1 - cos^2t = y^2/r^2 => cos^2t = 1 - y^2/r^2 = 1 - y^2/sec(2x) = 1 - y^2.cos(2x)

:. 1/cos(2x) = x^2/[1 - y^2.cos(2x)]
=> sec(2x) - y^2 = x^2
=> y^2 = sec(2x) - x^2
=> y = Sqrt[sec(2x) - x^2]

2.) x = rcost, y = rsint
r^2 = cosec(2x) = 1/(sin2x) = x^2/cos^2t

sint = y/r => 1 - cos^2t = y^2/r^2 => cos^2t = 1 - y^2/r^2 = 1 - y^2.sin(2x)

:. 1/(sin2x) = x^2/[1 - y^2.sin(2x)]
=> cosec(2x) - y^2 = x^2
=> y^2 = cosec(2x) - x^2
=> y = Sqrt[cosec(2x) - x^2]
Have i misunderstood the question? I was reading it as the polar co-ords were parametrised by (r,x) instead of (r,theta)
To finish off from my post use
x=rcosx y=rsinx ; x and y on LHS are the cartesian co-ordinates, the r and x are the polar co-ords.
If not then there is nothing complicated to do.
Since r^2=x^2+y^2
r^2 = sec(2x) gives x^2+y^2=sec2x
r^2 = cosec(2x) gives x^2+y^2=cosec(2x)
Reply 4
hmm for the first one I get something like (x^2 - y^2)^2 = x^2 + y^2, assuming the OP is refering to r = sec 2theta.
ssmoose
r^2 = sec(2x)

x^2 + y^2 = r^2
x = rsinx, y = rcosx

r^2 = sec(2x)
r^2.cos2x = 1
r^2(cos^2x - sin^2x) = 1
r^2(y^2/r^2 - x^2/r^2) = 1
y^2 - x^2 = 1
Reply 6
For the second one I get 2xy = 1.

r^2 = cosec(2t) = 1/sin(2t) = 1/(2cos(t)sin(t))
x/r = cos(t), y/r = sin(t).
r^2 = r^2/(2xy)

=> 2xy = 1.