How do you convert #x^2-y^2+25# into polar form?

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Answer 1 · 2018-02-17T18:17:33

One cannot really convert an expression into polar form; one needs an equation in one of the following forms:

#f(x,y) = c#, #x = f(y)#, or #y = f(x)#

because proper polar form is the radius, r, in terms of some function of #theta#:

#r = f(theta)#

The best that can be done, in this case, is to substitute #x = rcos(theta)# and #y = rsin(theta)#:

#(rcos(theta))^2-(rsin(theta))^2+25#

and then simplify:

#r^2(cos^2(theta)-sin^2(theta))+25#

However, this is not really polar form because, if one picks a value for #theta# on cannot determine the corresponding value(s) of r.