# Conic section equation

In polar coordinates, (1, 3/2) is $(\sqrt{13}/2, \theta)$ where $\theta = \arctan(1.5) \approx .98$ (radians). If you rotate this by $\pi/4$ you get $(\sqrt{13}/2, \theta + \pi/4)$, again in polar form. Can you work that out?
Ok I was finally able to draw this cursed parabola. May it go to hell. thank you again for your help! This page helped me a lot : https://en.wikipedia.org/wiki/Rotation_of_axes

In polar coordinates, (1, 3/2) is $(\sqrt{13}/2, \theta)$ where $\theta = \arctan(1.5) \approx .98$ (radians). If you rotate this by $\pi/4$ you get $(\sqrt{13}/2, \theta + \pi/4)$, again in polar form. Can you work that out?
ONE VERY LAST QUESTION. Is my equation in this form : y=(x^2)/2+1 in canonical form right ?

Mark44
Mentor
ONE VERY LAST QUESTION. Is my equation in this form : y=(x^2)/2+1 in canonical form right ?
I don't know what "canonical form" means.