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Maths problem involving Coordimate Geometry

  1. Mar 12, 2005 #1
    find the equation of a straight line whose x-intercept and y-intercept are a and b respectively.If this line varies such that [tex]\frac{1}{a^2}[/tex] + [tex]\frac{1}{b^2}[/tex] = [tex]\frac{1}{c^2} [/tex]with c as a constant,show that the locus of the foot of the perpendicular from the origin to this line is the curve
    [tex] X^2[/tex] +[tex] Y^2[/tex] = [tex]C^2 [/tex].


    I want to ask what is meant by the phrase 'foot of the perpendicular from the origin to this line ' ?

    I hope that somebody will help me to explain the meaning and thanks for anybody that spend some time on this question.
     
    Last edited: Mar 12, 2005
  2. jcsd
  3. Mar 12, 2005 #2
    The 'foot of the perpendicular from the origin to this line' means the curve which the intesection of the line and the perpendicular forms as [tex]a[/tex] and [tex]b[/tex] are varied. This intersection point changes with [tex]a[/tex] and [tex]b[/tex], and the path that it traces they move through all their possible values is what the question is looking for.
     
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