Finding asymptotes of polar equation r = tan2theta

[aznzunit]

I am trying to find the equations of the linear asymptotes of the polar equation r = tan 2theta.
I think i've already shown that the slopes of the asymptotes are 1 and -1 because in the equations for dy/dtheta and dx/dtheta, tan^theta - 1 is in the denominator. which means that when theta is equal to pi/4, the denominator is 0.

As to finding a y-intercept, the 4 asymptotes seem to intersect to create a square (assumption).
My initial approach was to consider the points on the rectangular plane (tan2theta * cos theta, tan2theta * sin theta) and ( tan2(90-theta) * cos(90-theta), tan2(90-theta) * sin (90-theta))
by finding the distance between these two points at infinity or lim as theta -> pi/4 of d^2 where d represents the distance between these two points, I think i could find the equations of the two lines, however it seems analytically that as theta approaches pi/4. But I seem to be finding that the distance between the two points is equal to zero.

Am i the only one getting that answer?

 

[tkhunny]

I'm worried with \(\displaystyle 2\theta\) that you are overlooking half the solution. I would want to investigate \(\displaystyle \theta=\frac{3\pi}{4}\) before I thought I was done. Note: It may challenge your working definition of an asymptote.

 

[aznzunit]

Can you clarify what you mean?
In rectangular form the equation is : x^6 - x^4 * y^2 - 4x^2y^2 - x^2y^4 + y^6 = 0.
it seems clear that the equation and its inverse are identical, so the line connecting point using theta and point using 90 - theta should be perpendicular to y = x.

 

[tkhunny]

Did you observe the Domain restrictions when converting to Cartesian Coordinates? If you noticed the x^2 - y^2 in the denominator along the way, you may also wish to included x = -y.