How do you convert #x^2-y^2=1# to polar form?

1 Answer
A. S. Adikesavan
Apr 28, 2016

#r^2=sec 2theta#.

Explanation:

#(x, y) = (r cos theta, r sin theta)#

So, #x^2-y^2=r^2(cos^2theta-sin^2theta)=r^2 cos 2theta=1#.

The semi-asymptotes are given by opposites #theta=pi/4 and theta=(5pi)/4# and, likewise, opposites #theta=(3pi)/4 and theta=(7pi)/4#.

In cartesian form, these equations are bundled to the simple

form #x+-y=0#, but the rotation effect given by #theta# is missing.. , .

Related questions
See all questions in The Trigonometric Form of Complex Numbers
Impact of this question
2735 views around the world
You can reuse this answer
Creative Commons License