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Math Help - parabola

  1. #1
    Member Veronica1999's Avatar
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    parabola

    Find an equation that says that P=(x,y) is equidistant from
    F= (2.0) and the y-axis. Plot four points that fit this equation.
    The configuration of all such points P is called a parabola.

    How come the equation (x-2)squared + y squared = xsquared + (y-y1)squared doesn't work?
    This does not seem to be a parabola.
    Last edited by Veronica1999; March 6th 2011 at 06:58 AM.
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  2. #2
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    You need to know one thing, that the shortest distance between a point and a line is the distance that is perpendicular to the line.

    Since \displaystyle (2, 0) is on the \displaystyle x axis, which is already perpendicular to the \displaystyle y axis, then the shortest distance from this point to the \displaystyle y axis is the distance along the \displaystyle x axis, i.e. \displaystyle 2 units.

    So that means that the distance between \displaystyle (x, y) and \displaystyle (2, 0) must also be \displaystyle 2 units.

    Therefore \displaystyle \sqrt{(x - 2)^2 + (y - 0)^2} = 2.

    Simplify this for the equation of your parabola.
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  3. #3
    Member Veronica1999's Avatar
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    I get (x-2)squared + ysquared = 4
    Isn't this a circle?
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  4. #4
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    Quote Originally Posted by Veronica1999 View Post
    Find an equation that says that P=(x,y) is equidistant from
    F= (2.0) and the y-axis. Plot four points that fit this equation.
    The configuration of all such points P is called a parabola.

    How come the equation (x-2)squared + y squared = xsquared + (y-y1)squared doesn't work?
    This does not seem to be a parabola.
    Let (x,y) be a point on the parabola. The distance to the focus is

    d^2=(x-2)^2+(y-0)^2

    Now the distance from the point (x,y) to the directrix is always measured perpendicular to the directrix. So in our case it is just the x coordinate. so

    d^2=x^2 equating the two gives

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