Convert the polar equation to rectangular form? r = 3/9 cos θ − 8 sin θ

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Answer 1 · 2016-11-30T21:47:09

Please see the explanation.

Multiply both sides by r:

$r^2 = 3/9rcos(theta) - 8rsin(theta)$

Substitute $x^2 + y^2$ for $r^2$ :

$x^2 + y^2 = 3/9rcos(theta) - 8rsin(theta)$

Substitute x for $rcos(theta)$ and y for $rsin(theta)$

$x^2 + y^2 = 1/3x - 8y$

Move the terms on the right to the left by adding $-1/3x + 8y$ to both sides of the equation:

$x^2 - 1/3x+ y^2 + 8y = 0$

We are going to complete the squares so add $h^2 + k^2$ to both sides of the equation:

$x^2 - 1/3x + h^2 + y^2 + 8y + k^2 = h^2 + k^2$

Set the middle term in the right side of the pattern $(x - h)^2 = x^2 -2hx + h^2$ equal to the corresponding term in the equation:

$-2hx = -1/3x$

Solve for h:

$h = 1/6$

Substitute the left side of the pattern into the equation:

$(x - h)^2 + y^2 + 8y + k^2 = h^2 + k^2$

Substitute $1/6$ for every h:

$(x - 1/6)^2 + y^2 + 8y + k^2 = (1/6)^2 + k^2$

Set the middle term in the right side of the pattern $(y - k)^2 = y^2 -2ky + k^2$ equal to the corresponding term in the equation:

$-2ky = 8y$

Solve for k:

$k = -4$

Substitute the left side of the pattern into the equation:

$(x - 1/6)^2 + (y - k)^2 = (1/6)^2 + k^2$

Substitute -4 for every k:

$(x - 1/6)^2 + (y - -4)^2 = (1/6)^2 + (-4)^2$

Combine the right side terms:

$(x - 1/6)^2 + (y - -4)^2 = 577/36$

Write the right side as a square:

$(x - 1/6)^2 + (y - -4)^2 = (sqrt(577)/6)^2$

Done!

This a circle with a radius of $sqrt(577)/6$ centered at $(1/6, -4)$