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Altered Parabolas (Posted on 2010-12-01) Difficulty: 3 of 5
Part 1.
Take a line and a point F not on the line. From a point on the line create a segment to F. Create the perpendicular bisector of this segment.

The envelope of these perpendicular bisectors using every point on the line is a parabola.

If we create perpendiculars that are some ratio, r, of the distance from the line to the point besides 1/2, what shape will the envelope become?

Part 2.

Take a line and a point F not on the line. Construct the set of all points equidistant from the point and line.

This set of points is a parabola.

If we find all the points that are d times the distance from the point as they are to the line, what shape will this set of points become?

This page indicates both constructions with a single interactive.

See The Solution Submitted by Jer    
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part 1: full solution | Comment 1 of 2

without loss of generality, let the line be the x-axis and the point be on the y-axis at (0,a), thus consider all points (t,0) on the line.  The point a ratio of r along the line is at (rt,ra) and the slope of the perpendicular is t/a, thus the family of lines under consideration is given by
y = ra + (t/a)*(x - rt)
y = ra + tx/a - rt^2/a
now, to find the nature of the boundary, check if there is a global maximum for y for a given x, thus solve dy/dt = 0
dy/dt = x/a - 2rt/a = 0
x/a = 2rt/a
t = x/(2r)
thus we get the equation
y = ra + x^2/(2ar) - x^2/(4ar)
y = ra + x^2/(4ar)
thus, the interesting result is that even if you change the ratio, you still get a parabola

  Posted by Daniel on 2010-12-04 02:58:13
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