Enlarging an ellipses along normal direction

Question

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

Answer 1

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 $a=b$.