How do you convert #r= tan θ# to rectangular form?

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Answer 1 · 2016-08-26T02:21:06

#y=+-x^2/sqrt(1-x^2), x in(-1, 1)#. The graph is through the origin and is symmetrical about the origin. As # x to +-1, y to +-oo#

Use the conversion formula #r(cos theta, sin theta) = (x, y)#.that givs

#r = sqrt(x^2+y^2), cos theta = x/r and sin theta = y/r#

Here, the conversion gives

#sqrt(x^2+y^2) =y/x# or explicitly, after solving for y,

#y =+-x^2/sqrt(1-x^2)# and for real #y, x in (-1. 1)#

Not to miss is my observation that the periodicity characteristic of

#tan theta, theta#, with period #pi#, for #theta in ( -oo, oo )# in polar

form #r = tan theta# is not explicit in the Cartesian form

#y =+-x^2/sqrt(1-x^2), x in (-1. 1)#