How do you write the given equation #x^2-y^2=1# into polar form?

1 Answer
Shwetank Mauria
May 30, 2017

#r^2=sec2theta#

Explanation:

Te relation between Cartesian coordinates #(x,y)# and polar coordinatess #(r,theta)# is given by

#x=rcostheta# and #y=rsintheta#

and hence #x^2-y^2=1# can be written as

#r^2cos^2theta-r^2sin^2theta=1#

or #r^2(cos^2theta-sin^2theta)=1#

or #r^2cos2theta=1#

or #r^2=sec2theta#

Note that when #theta# is between #45^@# and #135^@# as also between #225^@# and #315^@#, #sec2theta# is negative and hence in polar coordinates this curve does lie between these two regions.

graph{(x^2-y^2-1)(x+y)(x-y)=0 [-5, 5, -2.5, 2.5]}

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