What is the Cartesian form of #r = -csc^2theta+4sec^2theta #?

1 Answer
Shwetank Mauria
Jul 21, 2016

#x^2y^2=(4y^2-x^2)sqrt(x^2+y^2)#

Explanation:

Relation between polar coordinates #(r,theta)# and rectangular coordinates is given by #x=rcostheta#, #y=rsintheta#, #r^2=x^2+y^2# and #tantheta=y/x#.

Hence, #r=-csc^2theta+4sec^2theta#

#hArrr=-1/sin^2theta+4/cos^2theta# or

#r=-r^2/y^2+(4r^2)/x^2# or

#1=-r/y^2+(4r)/x^2=(r(-x^2+4y^2))/(x^2y^2)# or

#x^2y^2=(4y^2-x^2)sqrt(x^2+y^2)#

graph{x^2y^2=(4y^2-x^2)sqrt(x^2+y^2) [-25.1, 25.08, -12.55, 12.55]}

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