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Thread: Year 11 2u locus questions please help

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    Year 11 2u locus questions please help

    1. Find the expression got the square of the distance from the point P (x,y) to the point K(-1,3).
    2. Find an expression for the square of the distance of P(x,y) from the line y=0.
    3. The point P(x,y) moves so that it is equidistant from K and the line y=0. Show that the locus of P is a parabola and find its vertex

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    Re: Year 11 2u locus questions please help

    1. PK = sqrt[(x+1)^2 + (y-3)^2]
    square of the distance = PK^2 = (x+1)^2 + (y-3)^2

    2. Logically, distance of P from y=0 is always a vertical distance as y=0 is a horizontal line. Therefore, distance of P from the line y=0:
    |y-0| = |y|
    Square of the distance = y^2

    3. From the first two parts, you have the distance of PK and distance of P from y=0
    Equating the two:
    (x+1)^2 + (y-3)^2 = y^2
    (x+1)^2 + y^2 - 6y + 9 = y^2
    (x+1)^2 = 6y - 9
    (x+1)^2 = 6(y - 3/2)

    Therefore the locus of P is a parabola with vertex (-1, 3/2)
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