How do you convert $y^{2} = 4 (x + 1)$ $y^2= 4(x+1)$ into polar form?

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Answer 1 · 2017-09-11T19:55:17

see below

Use the formulas $x = r cos θ, y = r sin θ$ $x=rcos theta, y=r sin theta$

$y^{2} = 4 (x + 1)$ $y^2 = 4(x+1)$

$r^{2} {sin}^{2} θ = 4 (r cos θ + 1)$ $r^2sin^2 theta=4(rcos theta+1)$

$r^{2} {sin}^{2} θ = 4 r cos θ + 4$ $r^2 sin^2 theta=4rcos theta+4$

$:.r^2 sin^2 theta-4rcos theta-4=0$

How do you convert y^2= 4(x+1)y2=4(x+1)y^2= 4(x+1) into polar form?