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Math Help - parabola problem

  1. #1
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    parabola problem

    Ab is a chord of parabola y^2=4ax with its vertex at a.bc is drawn perpendicular to ab meeting the axis at c.the projection of bc on the axis of parabola is
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  2. #2
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    Quote Originally Posted by prasum View Post
    Ab is a chord of parabola y^2=4ax with its vertex at a.
    I don't understand what this means. "its" can only refer to the chord or the parabola. A chord does not have a vertex and the vertex of the parabola y^2= 4ax is at (0, 0).

    bc is drawn perpendicular to ab meeting the axis at c.the projection of bc on the axis of parabola is
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    the point 0,0 is A
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    So you mean "AB", not "Ab", "BC", not "bc", and the vertex is at "A" not "a". That's particularly confusing since you also used "a" as a number in the equation of the parabola.

    If B is a point on the parabola, then B= \left(\frac{y_0^2}{4a}, y_0\right) for some y_0 and the slope of AB is [tex]\frac{y}{\frac{y^2}{4a}}= \frac{4a}{y_0}[/quote]. Since BC is to be perpendicular to that, it must have slope -\frac{y_0}{4a} and it passes through B= \left(\frac{y_0^2}{4a}, y_0\right).

    Write the equation of that line and set y= 0 in it to find x in C= (x, 0). The "projection" of of BC on the x-axis is the segment from \left(\frac{y^2}{4a}, 0\right) to (x, 0).
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