solving conics

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Explain which conic section this equation and explain how to solve it:
12x^2-18y^2-18x-12y+12=0

  • solving conics -

    let's complete the square to get it into standard form

    12(x^2 - (3/2)x + .....) - 18(y^2 + (2/3)y + ...) = -12

    12(x^2 - (3/2)x + 9/16) - 18(y^2 + (2/3)y + 1/9) = -12 + 9/16 + 1/9

    12(x - 3/4)^2 - 18(y + 1/3)^2 = -1631/144
    divide by 1631/144
    (x-3/4)^2/(1728/1631) - (y+1/3)^2/(2592/1631) = -1

    I am sure you can determine the properties of the vertical hyperbola from there.
    Check my arithmetic.

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