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My friend sent me this problem, which (upon Googling) seems to be from a Cornell class (1220?). Anywho.

The ellipse

My advice to him was to parametrize the ellipse (say, in the first quadrant) with
$x = a \cos(t); y = b \sin(t)$, find normal lines, then use the formula for the distance between a point and a line. But then I started wondering:

Is there a better way? (Also, will this method even lead to a solution?!?)

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I can't see at once why your method won't work; at the very least, your parameter value can be constrained to lie within $[0,2\pi)$... –  J. M. Oct 25 '11 at 1:35
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It may (or may not) be easier to use the rational parameterization $t\mapsto(a\frac{1-t^2}{1+t^2},b\frac{2t}{1-t^2})$. Gets rid of the trigonometric functions. –  Henning Makholm Oct 25 '11 at 1:36
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@Henning: What can't the magic Weierstrass substitution solve, no? :) –  J. M. Oct 25 '11 at 1:47
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joriki: nice catch! –  The Chaz 2.0 Oct 25 '11 at 2:33
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Here's an animation from Wikipedia. –  Henning Makholm Oct 25 '11 at 15:28
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2 Answers

up vote 6 down vote accepted

The same maximal distance occurs in each quadrant, so we can restrict attention to $t\in[0,\pi/2]$. The tangent vector at $t$ is $(-a\sin t,b\cos t)$. This vector is normal to the line, so we just have to take the scalar product of a unit vector in this direction with the position vector in order to find the distance of the origin from the line:

$$ \begin{eqnarray} D &=& \left\lvert\frac{(-a\sin t,b\cos t)}{\sqrt{a^2\sin^2t+b^2\cos^2t}}\cdot(a\cos t,b\sin t)\right\rvert \\ &=& \frac{(a^2-b^2)\sin t\cos t}{\sqrt{a^2\sin^2t+b^2\cos^2t}}\;. \end{eqnarray}$$

Differentiating with respect to $t$ yields

$$\frac{a^2\sin^4 t-b^2\cos^4t}{\left(a^2\sin^2t+b^2\cos^2t\right)^{3/2}}\;,$$

and setting this to zero yields

$$a^2\sin^4t=b^2\cos^4t\;,$$

$$t=\arctan\sqrt{\frac ba}\;.$$

Using $\cos t=1/\sqrt{1+\tan^2 t}$, we can evaluate $D$ at this parameter:

$$ \begin{eqnarray} D &=& \frac{(a^2-b^2)\sin t\cos t}{\sqrt{a^2\sin^2t+b^2\cos^2t}} \\ &=& \frac{(a^2-b^2)\tan t}{\sqrt{a^2\tan^2t+b^2}}\cos t \\ &=& \frac{(a^2-b^2)\tan t}{\sqrt{a^2\tan^2t+b^2}}\frac1{\sqrt{1+\tan^2 t}} \\ &=& \frac{(a^2-b^2)\sqrt{b/a}}{\sqrt{a^2(b/a)+b^2}}\frac1{\sqrt{1+b/a}} \\ &=& \frac{a^2-b^2}{a+b}\;. \\ &=& a-b\;. \end{eqnarray}$$

The result obviously supports your idea that there might be a simpler way to do this.

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+1 Very nice. I esp. liked the part where you get the distance to the normal in one stroke using the position vector of the point. –  Srivatsan Oct 25 '11 at 2:12
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I wonder what your proof would look like if you translate to Weierstrass's form? –  J. M. Oct 25 '11 at 2:18
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Pushing the OP's suggestion.

  1. The tangent at the point $P = (a \cos t, b \sin t)$ is given by $b \cos t x + a \sin t y = ab$.

  2. The normal at $P$ is given by $a \sin t \ x - b \cos t \ y = (a^2-b^2)\sin t \cos t$.

  3. The distance of the origin from the normal is given by the formula: $$ D :=\frac{|(a^2 - b^2) \sin t \cos t|}{\sqrt{a^2 \sin^2 t + b^2 \cos^2 t}} = \frac{|a^2 - b^2|}{\sqrt{\frac{a^2}{\cos^2 t} + \frac{b^2}{\sin^2 t}}}. $$

  4. It remains to bound the distance $D$. By Cauchy-Schwarz, we have $$ \sqrt{(\cos^2 t + \sin^2 t)\left( \frac{a^2}{\cos^2 t} + \frac{b^2}{\sin^2 t} \right)} \geq (a+b). $$ Using the above inequality (and noting that $\cos^2 t + \sin^2 t = 1$), the distance can be upper bounded by $$ D \leq \frac{|a^2 - b^2|}{a+b} = |a-b|. $$ It is clear that equality is attained above when $\frac{a}{\cos^2 t} = \frac{b}{\sin^2 t}$; for instance, for $t = \arctan \sqrt{\frac{b}{a}}$ (focusing only on the positive quadrant).

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So $t = \arctan{\sqrt{\frac{b}{a}}}$? Great! –  The Chaz 2.0 Oct 25 '11 at 2:06
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Yes, as joriki mentions, that's the solution in the positive quadrant. There are other symmetric values of $t$ as well. EDIT: @TheChaz I have given only a sketch of the answers (I haven't shown how to calculate the equations of the tangent and normal, for e.g.) The details are straightforward to fill, so I hope this is ok. –  Srivatsan Oct 25 '11 at 2:07
    
You've done more than I did! I just told him what to do, while suspecting that the actual calculations might be tedious. –  The Chaz 2.0 Oct 25 '11 at 2:12
    
+1 -- nice idea to use Cauchy-Schwarz to get the maximum without differentiating. –  joriki Oct 25 '11 at 2:19
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