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Least distance from a point to a parabola

  1. Sep 10, 2007 #1
    1. The problem statement, all variables and given/known data
    There is a point (4,2), from which I have to find the least distance to the parabola y^2=8x. I.e, the least distance connecting the point and the parabola.

    2. Relevant equations

    Perpendicular distance = mod[(ax1+by1+c)/sqrt(a^2+b^2)]
    3. The attempt at a solution

    I already know the solution to this one. Our professor said the geometric way of solving this one may not seem easy. So we did this by using "Applications of Derivatives", i.e. Minima.

    Somehow, I found a geometrical way of solving this. It goes like this, if I take a point P(x1,y1) on the parabola such that, when a tangent is drawn to this point, a normal from the point of contact to this tangent passes through (4,2). By using the formula to find distance between a point and line, i.e. the point (4,2) and the tangential line yy1=2a(x+x1), and then using y1^2=2ax1, we can get a quadratic equation in x1 or y1, which one solving matches the given solution.

    I don't know how I landed on this, but why is the above true? The tangent I've considered is quite special, considering the normal to it from a point coincides with the point of contact. How does this deliver the least distance? Can someone shed some light on related facts?
  2. jcsd
  3. Sep 10, 2007 #2


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    Yes, you are correct. The "shortest distance" from a point to a curve is always along the line perpendicular to the curve, or perpendicular to the tangent line of the curve.

    Here, the curve is given by x= y2/8 so 1= y/4 dy/dx: dy/dx= 4/y. The slope of the normal line is then -y/4. So the shortest distance from (4,2) to that curve is along the line y= (-y0/4)(x-4)+ 2. That line intersects x= y2/8 when x= ((-y0/4)(x-4)+ 2)/8. You can solve that for x in terms of y0, then use the fact that x= y2/8 to find y0. Once you know that, you can find the point and so the distance.

    The reason that idea, that the shortest distance is along a perpendicular, works is that the hypotenuse is always the longest side in a right triangle. Drop a perpendicular from the given point ot the given line. Any other line from that point to that line is the hypotenuse of a right triangle and so is longer.

    That's for distance from a point to a line. For a curve, there may be several different points where there is a perpendicular from a point. All of those will be "local" minima but it may be that only one of them is the true minimum.
  4. Sep 10, 2007 #3
    The calc way is rather easy to minimize the distance between: y^2 = 8x and (4,2), and you should probably practice that too

    The distance between the two points can be found easily enough with the distance equation

    D = ((x_1 – x_2)^2 + (y_1 – y_2) ^2)^(1/2)

    So knowing that x=(1/8)y^2, and that the same point will minimize D and D^2 . Let’s let D* be the distance squared between the point and the parabola.

    We can find the distance with

    D* = ((1/8)y^2 – 4)^2 + (y – 2)^2)

    Now all you need to do is take the first derivative (with respect to y) of D* and set it = to 0, then solve. This will give you potential mins, so plug them into D* and see which one minimizes.
  5. Sep 10, 2007 #4
    Thanks for the explanation HallsofIvy, it made the things much more clearer than before.


    Yes, we've already followed that method. Though its quite short (and we need to solve such problems in under a minute and half or so), I still wanted to know the geometric position of what I was really doing.

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