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Math Help - Coordinates Question

  1. #1
    Member looi76's Avatar
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    Coordinates Question

    Question:
    The curve y^2 = 12x intersects the line 3y = 4x + 6 at two points. Find the distance between the two points.

    Attempt:

    y^2 = 12x
    y = (12x)^{\frac{1}{2}}

    3y = 4x + 6
    y = \frac{4}{3}x + 2

    (12x)^{\frac{1}{2}} = \frac{4}{3}x + 2
    12x = \left(\frac{4}{3}x + 2\right)^2

    12x = \left(\frac{4}{3}x + 2\right)\left(\frac{4}{3}x + 2\right)

    12x = \frac{16}{9}^2 + \frac{16}{3}x + 4

    = \frac{16}{9}^2 + \frac{16}{3}x - 12x + 4

    = \frac{16}{9}^2 - \frac{20}{3}x + 4

    a = \frac{16}{9} \ , \ b = -\frac{20}{3} \ , \ c = 4

    x = \frac{-b \pm \sqrt(b^2 - 4ac)}{2a}

    x = \frac{-(-\frac{20}{3}) \pm \sqrt((-\frac{20}{3})^2 - 4 \times \frac{16}{9} \times 4)}{2 \times \frac{16}{9}}

    Where did I go wrong?
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  2. #2
    Moo
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    Hi

    Quote Originally Posted by looi76 View Post
    Question:
    The curve y^2 = 12x intersects the line 3y = 4x + 6 at two points. Find the distance between the two points.

    Attempt:

    y^2 = 12x
    y = (12x)^{\frac{1}{2}}

    [...]
    Remember that if a^2=b^2, then a={\color{red}\pm} b

    ^^
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  3. #3
    Moo
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    For your problem, you have to solve simultaneously these equations :

    y^2=12x
    3y=4x+6

    Square the second equation

    9y^2=16x^2+48x+36

    Substitute the first equation into this one :

    9(12x)=16x^2+48x+36

    16x^2-60x+36=0 \longleftrightarrow 4x^2-15x+9=0

    4x^2-15x+9=(x-3)(4x-3)=0

    Therefore, x=3 or x=\frac 34

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