Curvature calculation of the ellipse at the end of its axes

Question

Calculate the curvature of the ellipse (of which the minor axis is b and the major axis is a) at the end of each axes!

Answer 1

Here's a geometric construction for the radius of curvature at point $P$ of an ellipse with foci $A$ and $B$.

• Construct the bisector $n$ of $\angle APB$, which is the normal at $P$.

• Construct on ray $n$ points $H$ and $K$, such that $PH$ and $PK$ are the diameters of the circles tangent to the ellipse at $P$ and passing through $A$ and $B$ respectively (note that $\angle PAH=\angle PBK=90°$).

• The radius of curvature is the harmonic mean of $PH$ and $PK$.

PROOF.

Let $P'$ be a point on the ellipse near to $P$ and let normal lines at $P$ and $P'$ meet at $C$. If we set $$ \alpha=\angle PAP',\quad \beta=\angle PBP',\quad \theta=\angle PCP', $$ then by angle chasing it is easy to find that $$ 2\theta=\alpha+\beta. $$ Consider now the circle through $APP'$, let $d_A$ be its diameter and $a$ its arc $PP'$ subtended by $\alpha$: we have then $a=\alpha d_A$. We can repeat the same reasoning for the circles through $BPP'$ and $CPP'$, with diameters $d_B$ and $d_C$, and arcs: $b=\beta d_B$, $c=\theta d_C$. Obtaining $\alpha$, $\beta$ and $\theta$ from these equalities, and plugging them into the above equation gives then: $$ 2{c\over d_C}={a\over d_A}+{b\over d_B}, $$ which can be rearranged as: $$ {1\over d_C}={1\over2}\left({a/c\over d_A}+{b/c\over d_B}\right). $$ In the limit $P'\to P$, the circles through $APP'$ and $BPP'$ tend to the circles tangent to the ellipse at $P$ and passing through $A$ and $B$ respectively, while $C$ tends to the center of the osculating circle at $P$. Hence $d_C$ tends to the radius of curvature $\rho$ at $P$, while $d_A$ and $d_B$ tend to $PH$ and $PK$ defined above. In addition, both $a/c$ and $b/c$ tend to $1$ as $P'\to P$. The above equation implies then: $$ {1\over \rho}={1\over2}\left({1\over PH}+{1\over PK}\right), $$ as it was to be proved.

The same argument can be repeated for a hyperbola and a parabola, leading to $$ {1\over \rho}={1\over2}\left|{1\over PH}-{1\over PK}\right| \quad\text{and}\quad \rho=2PH $$ respectively. A detailed explanation for the parabola can be found here.