How do you convert $r = - 4 cos (θ)$ $r=-4cos(theta)$ into cartesian form?

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How do you convert $r = - 4 cos (θ)$ $r=-4cos(theta)$ into cartesian form?

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Answer 1 · 2016-06-05T05:43:56

$x^{2} + y^{2} + 4 x = 0$ $x^2+y^2+4x=0$

Explanation:

To convert polar form of equation to Cartesian form, one can use $r cos θ = x$ $rcostheta=x$ and $r sin θ = y$ $rsintheta=y$ . From this we also get $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ .

Hence, $r = - 4 cos θ$ $r=-4costheta$ can be written as

$r^{2} = - 4 r cos θ$ $r^2=-4rcostheta$ or

$x^{2} + y^{2} = - 4 x$ $x^2+y^2=-4x$ or

$x^{2} + y^{2} + 4 x = 0$ $x^2+y^2+4x=0$

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How do you convert $r = - 4 cos (θ)$ $r=-4cos(theta)$ into cartesian form?

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How do you convert r=-4cos(theta)r=−4cos(θ)r=-4cos(theta) into cartesian form?

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How do you convert $r = - 4 cos (θ)$ $r=-4cos(theta)$ into cartesian form?