Andre W. answered • 09/25/13

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To bring a conic section equation (in this case, an ellipse) into vertex form, you need to complete the squares on x and y and make the right side equal to 1. Here are the steps:

(3x²-36x) + (4y²+24y) + 132 =0

3(x²-12x) + 4(y²+6y) + 132 =0

3(x-6)²-108 + 4(y+3)²-36 + 132=0

3(x-6)² + 4(y+3)² = 12

(x-6)²/4 + (y+3)²/3 = 1

Thus, the ellipse is centered at (6,-3). To find its axes and vertices, re-write the equation as

((x-6)/2)² + ((y+3)/√3)² = 1

Thus, the semimajor and semiminor axes are 2 and √3, respectively. The four vertices are at

(6±2,-3) and (6,-3±√3).