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

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Answer 1 · 2016-02-15T07:32:12

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

from the given, $r = - 4 cos θ$ $r=-4 cos theta$

multiplying both sides of the equation by $r$ $r$

$r \cdot r = r \cdot (- 4 cos θ)$ $r*r=r*(-4 cos theta)$
$r^{2} = - 4 \cdot r cos θ$ $r^2=-4*r cos theta$

Using $r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$ and $x = r cos θ$ $x=r cos theta$

it follows

$r^{2} = - 4 \cdot r cos θ$ $r^2=-4*r cos theta$

${(\sqrt{x^{2} + y^{2}})}^{2} = - 4 \cdot x$ $(sqrt(x^2+y^2))^2=-4*x$

$x^{2} + y^{2} = - 4 \cdot x$ $x^2+y^2=-4*x$

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

have a nice day, ... from the Philippines !

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