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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 2: full solution Comment 2 of 2 |

again, let the line be the x-axis and the point be (0,a)
now for any point (x,y), the distance to the line is y and the distance to the point is sqrt(x^2+(y-a)^2)
thus we want
d*y = sqrt(x^2 + (y-a)^2)
d^2y^2 = x^2 + y^2 - 2ay + a^2
x^2 + (1-d^2)y^2 - 2ay + a^2 = 0
thus if d<1 then we have an ellipse
if d=1, we have a parabola
if d>1, then we have a hyperbola

what is interesting about this one, is that you can only get 3 out of the 4 conic sections.  There are no values of d,a that will give you a circle, unless you consider the degenerative case of d=0, in which case you get the point (0,a) and it is arguable if that should be considered a circle.

  Posted by Daniel on 2010-12-04 03:11:54
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