Question

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

Answer 1

`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]}