How do you convert $r = \frac{6}{2 - 3 sin (θ)}$ $r=6/(2-3sin(theta))$ to rectangular form?

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Answer 1 · 2016-06-19T18:58:50

$4 x^{2} - 5 y^{2} - 36 y = 36$ $4x^2 -5y^2 -36y =36$

Simplify the expression by cross multiplication and write $2 r = 6 + 3 r sin θ$ $2r=6+ 3rsin theta$

Or, $2 r = 6 + 3 y [r sin θ = y]$ $2r= 6+3y [ rsin theta=y]$

Now square both sides to have $4 r^{2} = {(6 + 3 y)}^{2}$ $4r^2= (6+3y)^2$

$4 (x^{2} + y^{2}) = 36 + 36 y + 9 y^{2}$ $4( x^2 +y^2)= 36 +36y +9y^2$

$4 x^{2} - 5 y^{2} - 36 y = 36$ $4x^2 -5y^2 -36y =36$

How do you convert r=6/(2-3sin(theta))r=62−3sin(θ)r=6/(2-3sin(theta)) to rectangular form?