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What is the Cartesian form of #(r-1)^2 = tan^2theta-4sectheta #?

1 Answer

A. S. Adikesavan

Feb 18, 2017

#x^2(x^2+y^2)+(x^2-y^2)+2xsqrt(x^2+y^2)(2-x)=0#. See Socratic graph, for the Cartesian form.

Explanation:

Use #r(costheta, sintheta)=(x, y)#, giving #r=sqrt(x^2+y^2)>=0#.

.Here,

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

Expansion and simplification give the answer.
graph{x^2+y^2+sqrt(x^2+y^2)(4/x-2)+1-y^2/x^2=0 [-10, 10, -5, 5]}

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