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The problem is as follows: Find the point on the parabola $2x=y^{2}$ closest to $(1,0$).

I was highly surprised because I ended up with the correct answer doing something completely different than the solutions manual suggested, and now I am wondering if my method is correct, or if it was simply due to pure luck.

This is how I solved it:

The point in question is (1,0), which means that the parabolic point is in the first quadrant. This way I can avoid implicit differentiation and do the following:

$y=\sqrt{2x} \implies y '(x)= (\sqrt{2x})^{-1} \implies slope\:of\:normal=-\sqrt{2x}$

My logic: The normal of the parabola that crosses (1,0) must also cross the point on the parabola closest to (1,0). Is this logic correct?

Using the point-slope equation to construct the normal line yields:

$y=-x\sqrt{2x}+\sqrt{2x}$

The point is found where the parabola and the normal intersects:

$\sqrt{2x}=-x\sqrt{2x}+\sqrt{2x} \implies x=0 \implies y=0$.

Is this a legitimate method of solving the problem?

DeepSea
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Akitirija
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  • You have a problem with you notation. You use $x$ for the point you are looking for and $x$ for the equation of the line. Just call $(x_0,y_0)$ the point on the parabola and if you follow the same steps you will discover the mistake. – mfl Oct 08 '14 at 21:22
  • Ah, too bad. Thank you mfl, I will definitely try to stick to your advice from now on to save some time and disappointment. Thanks for your advice! – Akitirija Oct 08 '14 at 21:25
  • You're welcome. – mfl Oct 08 '14 at 21:26
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    I want to add, though, that the basic idea behind the method -- that the normal at the closest (or farthest) point must pass through the target -- is sound, at least under reasonable assumptions. Only the execution was a little flawed. (There's also the possibility that the closest point is one endpoint of the interval under consideration, but that's a small detail). – John Hughes Oct 08 '14 at 21:29

3 Answers3

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Let $(x,y) = \left(\dfrac{y^2}{2},y\right)$ be the point on the parabola that is closest to $(1,0)$, then this point is where the distance of the two point is minimized. $d = \sqrt{\left(\dfrac{y^2}{2} - 1\right)^2 + y^2} = \sqrt{\dfrac{y^4}{4} + 1} \geq 1$, and is minimized when $y = 0$. Thus the sought point is the origin $(0,0)$.

DeepSea
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  • To be honest, I do not understand the formula (d=...). Is this due to the fact that I do not know parabolas? If this is the case, I will read more about parabolic and hyperbolic functions. – Akitirija Oct 08 '14 at 21:48
  • The formula is just the distance between $({ 1 \over 2} y^2, y)$ and $(1,0)$??? – copper.hat Oct 08 '14 at 21:53
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notice that the squared distance from $(x,y)$ on the curve to $(1,0)$ is $$ a^2 = (x-1)^2 +y^2 = x^2 +1 $$ from this it is immediate from this where the minimum occurs

David Holden
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Your problem can be expressed as $\min_t \|p-f(t)\|^2$, where $p = (1,0)$ and $f(t) = ({1 \over 2} t^2, t)$. At as solution we will have $-(p-f(t))^T f'(t) = 0$, or in other words, the vector $p-f(t)$ is perpendicular to $f'(t)$.

In your case $f'(t) =(t,1)$, so we are looking for solutions $({1 \over 2} t^2 -1, t)^T(t,1) = 0$. This gives ${1 \over 2} t^3-t+t = 0$ from which it follows that $t=0$ and hence the solution is $(0,0)$.

copper.hat
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  • Thank you for your answer, but I fear that this is way over my head for now, unfortunately. – Akitirija Oct 08 '14 at 21:53
  • Well, it is justifying your approach. The difference is that you used the parametrization $(t, \sqrt{2t})$ of the upper half of the parabola whereas I used $({1 \over 2}t^2, t)$ which parametrizes the entire parabola. – copper.hat Oct 08 '14 at 21:56