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

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Answer 1 · 2017-02-18T13:09:50

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

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]}

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