How do you convert # (r+1)^2= theta + sectheta # to Cartesian form?

1 Answer
Shwetank Mauria
Jun 29, 2016

#x/y=tan[x^2+y^2+sqrt(x^2+y^2)(2-1/x)+1]#

Explanation:

When we convert polar coordinates #(r,theta)# to Cartesian coordinates, the relation is #x=rcostheta#, #y=rsintheta# and hence #r^2=x^2+y^2# and #theta=tan^(-1)(x/y)#.

Hence #(r+1)^2=theta+sectheta# can be written as

#r^2+2r+1=theta+sectheta# or

#x^2+y^2+2sqrt(x^2+y^2)+1=tan(-1)(x/y)+sqrt(x^2+y^2)/x# or

#x^2+y^2+sqrt(x^2+y^2)(2-1/x)+1=tan(-1)(x/y)# or

#x/y=tan[x^2+y^2+sqrt(x^2+y^2)(2-1/x)+1]#

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