PM is a segment of ellipse normal upto intersecting point $M$ on y- axis and PC is the radius of curvature.
EDIT1:
Show that (segment lengths of) PC is proportional to the cube of PM, the constant of proportionality being $ (b^2/a^4)$ at all points $P$. Two locations are verified as under:
At $a$:
$$ \frac{PC}{PM^3}=\frac{b^2/a}{a^3}=\frac{b^2}{a^4} $$
At $b:$
$$ \frac{PC}{PM^3}=\frac{a^2/b}{(a^2/b)^3}= \frac{b^2}{a^4} $$
Take the standard parameterization of the ellipse:
$$P := (a\cos t, b\sin t) \tag{1}$$ where $a$ and $b$ are the "horizontal" and "vertical" radii, not necessarily "major" and "minor". (The major/minor distinction is immaterial.) To avoid sign complications, we'll consider the first-quadrant arc of the ellipse, where $0\leq t\leq \pi/2$. With this, we are assured that an "inward-pointing" normal at $P$ is given by $$n := (-b \cos t, -a \sin t) \tag{2}$$ (which is obtained from exchanging the components of the tangent vector $P'(t)$, and changing signs to ensure the proper orientation). A point $K$ at distance $k$ from $P$ along the normal line has the form $$K := P + \frac{k}{|n|} n = \left(\; \left(a-\frac{bk}{|n|}\right) \cos t,\;\left(b-\frac{ak}{|n|}\right)\sin t\;\right) \tag{3}$$ In particular, setting appropriate coordinates to zero, we find that the point $X$ and $Y$ on the $x$- and $y$-axes corresponds to the distances $$|PX| =\frac{b}{a}|n| \qquad |PY| = \frac{a}{b}|n| \tag{4}$$ Now, the point $Z$ on the evolute has distance from $P$ equal to the radius of curvature of the ellipse at $P$. By the parametric formula, we have
$$|PZ| := \frac{\left(P_x'^2 + P_y'^2\right)^{3/2}}{\left|P_x'' P_y'-P_x'P_y''\right|} = \frac{|n|^3}{a b} \tag{5}$$
(where I'm using $P_x$ and $P_y$ to refer to the coordinates of $P$). Thus,
and the result follows. $\square$