What is the limit of the y-intercept as P approaches O on a given parabola?

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SUMMARY

The limit of the y-intercept, denoted as \( b \), approaches \( \frac{1}{2} \) as point \( P \) approaches the origin \( O \) on the parabola defined by the equation \( y = ax^2 \). This phenomenon occurs due to the geometric properties of the perpendicular bisector of segment \( OP \). As \( P \) moves closer to \( O \), the slope of the line connecting \( O \) and \( P \) influences the position of the y-intercept, leading to this surprising limit.

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  • Understanding of parabolic equations, specifically \( y = ax^2 \)
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We are given the parabola $y \,=\,ax^2$
. . It opens upward, is symmetric to the y-axis, with vertex at the origin $O$.

Select any point $P(p,ap^2)$ on the parabola.

Construct the perpendicular bisector of $OP$
. . and consider its $y$-intercept, $b.$

Code:
                  |
                 b|
     ◊            ♥            ◊
                  |\
                  | \             P
      ◊           |  \        ♠(p,ap^2)
                  |   \     *
       ◊          |    \  *  ◊
        ◊         |     *   ◊
          ◊       |   *   ◊
             ◊    | *  ◊
    - - - - - - - ◊ - - - - - -
                  |O

Find $\displaystyle\lim_{P\to O}b$

The answer is surprising.
Can anyone explain this phenomenon?
 
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(Hide the spoiler from the forum overview.)
The y-intercept is at twice the distance to the focal point.

This is similar to a lens.
If you have a point source at twice the focal distance of a lens, the light rays converge at the other side at twice the focal distance.

In this case we have a parabolic mirror.
When light rays start at twice the focal distance, they return to the same point.
 

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