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Maxima and Minima question

  1. Feb 2, 2013 #1
    A point P whose x-coordinates is a is taken on the line y=3x-7. If Q is the point(4,1)
    show that PQ2 = 10a2-56a+80. Find the value of a which will
    make the expression a minimum. Hence show that the coordinates of N, the foot of the perpendicular from Q to the line are (24/5 , 1 2/5). Find the equation of QN.

    I'm having trouble with the first part

    isn't P {a, (3a-7)} and Q2 ( 16,1)?
  2. jcsd
  3. Feb 2, 2013 #2

    Ray Vickson

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    No. ##PQ^2## is the square of the distance from P to Q; it should have been written as ##(PQ)^2## or maybe as ##\vec{PQ}^2##.
  4. Feb 2, 2013 #3
    Oh okay thanks for the 2nd part that value that makes the expression a minimum is it 2.8? and is Q on the curve? cause to do the last part I think I need the equation of the tangent at Q.
  5. Feb 2, 2013 #4

    Ray Vickson

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    You need to show your work.
  6. Feb 2, 2013 #5
    PQ2 = 10a2 - 56a +80

    dy/dx = 20a -56

    a= 2.8
  7. Feb 2, 2013 #6
    Q is not on the line, but N is. In fact, it's the orthogonal projection of Q into the line. The value of a you obtained is associated to the minimal distance between the line and the point Q, which is itself associated with the concept of an orthogonal projection. If you plug in 2.8 into (a, 3a - 7), you get precisely the point N, with the coordinates they mention. Having Q and N, finding the equation of the line QN should be straightforward.
  8. Feb 2, 2013 #7
    Oh thank you!
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