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Show that a locus of points is a hyperbola

Consider the set of points P such that the distance from P to the point (2,3) is equal to the sum of the distances from P to the two coordinate axes.

  1. Show that the part of this set of points lying in the first quadrant forms a hyperbola. Locate the asymptotes of this hyperbola and make a sketch.
  2. Sketch the set of points in the other quadrants.

Incomplete.

2 comments

  1. S says:

    The basic equation is (x-2)^2 + (y-3)^2 = |x+y|. If we suppose there is a hyperbola with the focus (2, 3), we can prove there is indeed a hyperbola with F=(2, 3), N_directrix=(1/sqrt2, 1/sqrt2), e=sqrt(2). As far as I can see, the hyperbola actually fits the locus in all quadrants. To get the asymptotes, just let first x tend to infinity, and then y tend to infinity…To get the sketch, it’s quite a tedious process, but one can get the focus in the Q3, by knowing the distances from the directrix to the focus in the Q1.

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