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]}

Related questions
See all questions in Converting Between Systems
Impact of this question
1545 views around the world
You can reuse this answer
Creative Commons License