Given an ellipses, enlarge it along normal direction a fixed length say 1cm. Do we get another ellipses? If so, how to prove ?


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  • 1
    $\begingroup$ You don't usually get an ellipse: for example, consider a very long and skinny ellipse, so skinny that it looks like a line segment. Then if you enlarge it along the normal direction, you will get a shape that looks like a rectangle with semicircles stuck onto two of the edges. $\endgroup$ – zeb Jun 20 '13 at 8:54
  • 4
    $\begingroup$ I think you never get another ellipse except in the trivial case where the original ellipse is a circle. It's a curve of degree four. Cauchy and Cayley looked at this: link.springer.com/article/10.1007%2FBF03198048 $\endgroup$ – alvarezpaiva Jun 20 '13 at 10:06
  • $\begingroup$ This is more difficult than I thought (deleted an answer) because the expanded ellipse may have different foci than the original. $\endgroup$ – coffeemath Jun 25 '13 at 16:26

Bring the ellipse to standard form $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a\ge b\tag1$$ The new curve is also symmetric in $x$- and $y$-axes. Its $x$-intercepts are $\pm (a+1)$, the $y$-intercepts are $\pm (b+1)$. If it were an ellipse, its equation would be $$\frac{x^2}{(a+1)^2}+\frac{y^2}{(b+1)^2}=1 \tag2$$ Now recall two standard facts:

  1. The radii of curvature of the ellipse (1) at the vertices are $b^2/a$ and $a^2/b$. (Source)
  2. For a convex curve, the radius of curvature of parallel curve at outer distance $d$ is $R+d$ where $R$ is the radius for the original curve at the nearest point. This is proved here and is geometrically evident by taking the Minkowski sum of convex set with the disk of radius $d$.

Combining 1 and 2, we conclude that $$ \frac{a^2}{b}+1 = \frac{(a+1)^2}{b+1} \tag3 $$ But the difference of two sides in (3) is $(a-b)^2/(b^2+b)$. Thus the parallel curve is an ellipse only when $a=b$.


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