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I am supposed to solve these conics, can you please help :)I am supposed to name the center and the vertices.

1. 3y^2-2x^2+12x+24+24y=0

  • Pre-Calculus -

    complete square again and again

    3 y^2 + 24 y = 2 x^2 -12 x - 24
    divide by 3
    y^2 + 8 y = (2/3) x^2 - 4 x - 8
    add square of half of 8
    y^2 + 8 y + 16 = (2/3) x^2 - 4 x + 8
    (y+4)^2 = -------
    now multiply by (3/2)
    (3/2)(y+4)^2 = x^2 - 6 x + 12
    move 12 over
    (3/2)(y+4)^2 -12 = x^2 - 6 x
    add square of half of 6
    (3/2)(y+4)^2 -3 = x^2 - 6 x + 9
    (3/2)(y+4)^2 -3 = (x-3)^2
    (3/2)(y+4)^2 - (x-3)^2 = 3

    (y+4)^2 /2 - (x-3)^2/3 = 1

    Patience, patience :)

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