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ANSWER

b.

EXPLANATION

The given polar equation is,

[tex]r = \frac{12}{6 + 4 \cos( \theta) } [/tex]

Cross multiply to get,

[tex]r(6 + 4 \cos( \theta) ) = 12[/tex]

[tex]6r + 4r\cos( \theta) ) = 12[/tex]

Substitute,

[tex] r\cos( \theta) = x[/tex]

and

[tex]r = \sqrt{ {x}^{2} + {y}^{2} } [/tex]

[tex]6 \sqrt{ {x}^{2} + {y}^{2} } + 4x = 12[/tex]

[tex]3\sqrt{ {x}^{2} + {y}^{2} } + 2x = 6[/tex]

[tex]3\sqrt{ {x}^{2} + {y}^{2} } = 4 - 2x [/tex]

[tex]9({x}^{2} + {y}^{2}) =( 4 - 2x ) ^{2} [/tex]

[tex]9{x}^{2} + 9 {y}^{2} =16 - 16x + 4x ^{2} [/tex]

[tex]5{x}^{2} + 9 {y}^{2} + 16x = 16[/tex]

This is an ellipse whose center is shifted to the left with orientation on the x-axis

The correct choice is B.
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