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#