Re: Distance between a point and y = ax^2 + bx + c


Arturo Magidin wrote:
In article <20061031234246.931$lw@xxxxxxxxxxxxxx>,
David W. Cantrell <DWCantrell@xxxxxxxxxxx> wrote:
magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:
In article
<1235297.1162325667272.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>, Magnus
<maol9883@xxxxxxxxxxxxx> wrote:
[snip]
I'm trying arbitrary functions y = ax^2 + bx + c with arbitary points
to find the closest point between.

Given a point (x0,y0), some parabolas have a unique "closest point" to
it. Some have more than 1; for example, the closest points to (0,1) on
the parabola x=y^2 are (0,0), (1,1), and (-1,1), all of them distance
1.

Three points?! There are at most _two_ closest points.

Quite right. My mistake there.

Just to finish the calculation, minimum distance from (0,1)
to y = x^2 occurs at ( 1/sqrt(2), 1/2 ) with local maximum
at (0,0). Minimum distance is thus sqrt(3)/2.

However if the point were (0,1/2), the minimum distance
would be attained at the vertex as a triple root of the
critical equation. Any closer to the vertex (along the axis
of symmetry) and one has a simple root at the vertex
and two (purely) imaginary roots.

--c

.

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