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So I have this problem for my homework:

Consider the ellipse: $\dfrac {x^2}{a^2} + \dfrac{y^2}{b^2}=1$ where $0. For every point $(x, y)$ on the ellipse find the the perpendicular line to the ellipse so that the point $(x, y)$ is on that line. This line cuts the ellipse in another point $(x', y')$. Prove that the distance between these two points is $$D(x, y)=\dfrac {2\left(\dfrac {x^2}{a^4} + \dfrac{y^2}{b^4}\right)^{3/2}}{\dfrac {x^2}{a^6} + \dfrac{y^2}{b^6}}\ .$$

So I've already done that; but after that it says:

Use Lagrange Multipliers to minimize the function $D(x, y)$.

But the equations get complicated and messy and I don't know if I have to consider the line or just the ellipse. Logically the minimum is at $(0, b)$. Can somebody help me?

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    but 0 isn't in the elipse, the profersor wants ou to find the pont (x,y) on the elipse such that the distance to (x',y') is minumun2012-12-03
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    Doing a quick experiment using [Cinderella](http://www.cinderella.de/), it seems that your intuition about what minimum point you expect is wrong: there are points off the axes of symmetry where the distance becomes a lot smaller. The larger the excentircity, the more pronounced this effect becomes.2012-12-03
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    http://en.wikipedia.org/wiki/Lagrange_multipliers describes how to maximize $f$ subject to $g=c$. So you'd have $g=c$ as the formula of your ellipsis, and $-D$ as the thing you want to maximize. Now plug these things into the formulas on Wikipedia, compute three derivatives, and you should have a set of three equations. Are you allowed to do this using a computer algebra system? If not, do things become messy enough to actually prevent a manual solution? If so, then re-examine your term above (which I have not checked) and see whether there might be some error in there.2012-12-03

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