What is the Cartesian form of $r = -sintheta+4cos^2theta$ ?

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Answer 1 · 2016-01-26T07:20:08

$(x^2+y^2+y)sqrt(x^2+y^2)=4x^2$

graph{(x^2+y^2+y)sqrt(x^2+y^2)=4x^2 [-8.046, 8.046, -4.02, 4.024]}

From the given

$r=-sin theta+4*cos^2 theta$

the conversion equations:

$r=sqrt(x^2+y^2)$ and $theta=tan^-1(y/x)$

$tan theta=y/x$ and $sin theta=y/sqrt(x^2+y^2)$ and $cos theta=x/sqrt(x^2+y^2)$

Convert now, the given equation becomes

$sqrt(x^2+y^2)=-y/sqrt(x^2+y^2)+(4*x^2)/(x^2+y^2)$

multiply both sides of the equation by $(x^2+y^2)$

$(x^2+y^2)sqrt(x^2+y^2)=-(y*(x^2+y^2))/sqrt(x^2+y^2)+((x^2+y^2)(4*x^2))/(x^2+y^2)$

$(x^2+y^2)sqrt(x^2+y^2)=-(ycancel((x^2+y^2)))/cancelsqrt(x^2+y^2)+(cancel((x^2+y^2))(4*x^2))/cancel(x^2+y^2)$

$(x^2+y^2)sqrt(x^2+y^2)=-y*sqrt(x^2+y^2)+4*x^2$

$(x^2+y^2)sqrt(x^2+y^2)+y*sqrt(x^2+y^2)=4*x^2$

$(x^2+y^2+y)*sqrt(x^2+y^2)=4*x^2$

What is the Cartesian form of r = -sintheta+4cos^2theta r = -sintheta+4cos^2theta ?