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Find the point(s) on the parabola x= y^2-8y+18 closest to (-2,4)

  1. May 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the point(s) on the parabola x= y^2-8y+18 closest to (-2,4)

    2. Relevant equations

    3. The attempt at a solution
    Using the distance formula I found that d^2 = y^4 - 16y^3 + 105y^2 - 328y + 416
    after differentiating I have 2d' = 4y^3 - 48y^2 + 210y -328
    I'm not sure what to do now, or what I may have done wrong, any help is appreciated.
  2. jcsd
  3. May 5, 2013 #2
    I'd have used the fact that the point closest to (-2,4) will lie on the line that passes through (-2,4) and is perpendicular to the parabola at the point where they cross.
  4. May 5, 2013 #3
    I don't understand what you mean. We were just taught to use the distance formula D^2 = (X2-X1)^2 + (Y2-Y1)2
    I subbed in -2 as x1 and 4 as y1 then the equation of x= y^2 -8y + 18 as X2. Am I doing it wrong?
  5. May 5, 2013 #4
    I feel like using the distance formula is over complicating things. After all, you haven't been asked to find the ACTUAL distance, just the point which is closest. You could do that by minimising the distance, but I think it's easier to think about what this would look like on the graph.

    So, the closest point on the parabola is the point which lines on the line which passes through (-2, 4) and intersects the parabola at a right angle.
  6. May 5, 2013 #5
    If you have studied Lagrange's multipliers, that might be the easiest solution.

    Otherwise, stick with Unredeemed's advice about the geometrical consideration: minimizing the metric formula yields a cubic equation, as you have already learnt.
  7. May 5, 2013 #6
    Unredeemed, can you please show me how to go on with the process you are talking about? We were taught to find the distance formula then differentiate and solve when it is 0. I've never studies Lagrange's multipliers.
  8. May 5, 2013 #7
    Voko, what would the constraint equation be if you applied the method of Lagrange multipliers?
  9. May 6, 2013 #8

    Ray Vickson

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    So far, so good. Now you just need to solve a cubic equation in y. You could try using the Rational Root Theorem, or try to solve the equation numerically, or use the exact formula for the roots of a cubic.
  10. May 6, 2013 #9


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    You don't have to use any Calculus for this particular problem. Completing the square, [tex]x= y^2-8y+18= y^2- 8y+ 16+ 2= (y- 4)^2+ 2. Notice what the vertex is.
  11. May 6, 2013 #10
    The equation of the parabola. Minimizing the square of the distance to the point given.
  12. May 6, 2013 #11


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    Follow HallsofIvy's hint, write the equation of the parabola in the form x=(y-4)^2+2, by completing the square. Write up d^2 with that formula without expanding it and differentiate. You can also plot out the function and the point and see the solution at once

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