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13 November, 01:41

4x^2+4y^2+36y+5=0 how do I put it in standard form

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  1. 13 November, 04:59
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    4x^2+4y^2+36y+5=0

    ... x^2 and y^2 with equal coefficient - > circle with standard form (x-h) ^2 + (y-k) ^2 = r^2

    ... group x, y and constant

    (4x^2) + (4y^2 + 36y) = - 5

    ... eliminate x, y coefficient (div by 4)

    x^2 + (y^2 + 9y) = - 5/4

    ... add necessary term to form

    x^2 + 2ax + a^2 = (x + a) ^2

    x^2 + [y^2 + 2 (9/2) y] = - 5/4

    ... need (9/2) ^2 for y-term

    x^2 + [y^2 + 2 (9/2) y + (9/2) ^2] = - 5/4 + (9/2) ^2

    x^2 + (y + 9/2) ^2 = 19 = (sqrt (19)) ^2
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