How do you convert $r = (8secθ) / (4secθ + 1)$ into rectangular form?

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Answer 1 · 2016-06-15T11:35:51

$15x^2+16y^2+16x-64=0$

We know that if a point in X-Y
plane has rectangular coordinate $(x,y)$ and its polar coordinate is $(r,theta)$ ,then we have the relation

$x=rcostheta and y =rsintheta$

and $r=sqrt(x^2+y^2)$

Now the given equation is

$r=(8sectheta)/(4sectheta+1)$

$=>r=((8sectheta)*costheta)/((4sectheta+1)*costheta)$

$=>r=8/(4+costheta)$

$=>4r+rcostheta=8$

$=>4sqrt(x^2+y^2)+x=8$

$=>(4sqrt(x^2+y^2))^2=(8-x)^2$

$=>16x^2+16y^2=64-16x+x^2$

$=>15x^2+16y^2+16x-64=0$

How do you convert r = (8secθ) / (4secθ + 1)r = (8secθ) / (4secθ + 1) into rectangular form?