Area of an ellipse proportional to integral of cross-ellipse distances?

Question

I am curious if the area of an ellipse can be shown to be proportional to the integral of all cross-ellipse distances. Before I define cross-ellipse distance, I will give a motivating example from a circle.

Suppose you have a circle with diameter $d$. Then $A_\text{circle}=\pi d^2/4$, and $C=\pi d$ . Now consider a line tangent to the edge of the circle that touches a point $p$ on the edge. The maximal distance from $p$ to any other point on the edge of the circle is equal to $d$. For a given $p$, call such a distance $e(p)$. If I integrate over all these distances, $e_\text{circle}(p)$, I get (perhaps erroneously) \begin{equation} \int e_\text{circle}(p) \, dp = Cd = \pi d^2 = 4A_\text{circle} \end{equation} Here, this seems simple since the maximal distance of any $p$ to another point on the hull of the circle is always $d$.

I am curious about whether something regarding the proportionality of this sort of distance to the area can be said for an ellipse.

Given a point $p$ on the hull of an ellipse, the maximum distance between $p$ and any other point on the hull is to the point $p_\text{far}$ such that a line passing through $p$ and $p_v{far}$ is orthogonal to the tangent touching $p$. Denote such a distance $e_\text{ellipse}(p)$. On a circle $e(p)=d$ but here it varies with $p$.

The area of an ellipse is $A_\text{ellipse} = \pi a b$, where $a$ and $b$ are the semi-major and semi-minor axes ($\frac 1 2$ of the ellipse's major and minor axes), respectively.

Is \begin{equation} \int e_\text{ellipse}(p) \, dp \propto A_\text{ellipse} \end{equation}

Can I show this by using linear combinations of $a$ and $b$ to characterize the distances $e_\text{ellipse}(p)$?

If it is true that the integral of the $e(p)$ is proportional to the area for any convex shape, I would really appreciate an argument or source.

In case it helps to give context, I am interested in this approach as a way to show that for two ellipses with $a_1, b_1$ for the first and $a_2, b_2$ for the second that using this approach can show that the area of one of the ellipses is larger than the area of the other ellipse.

Answer 1

No, this is not true. Take the ellipse with semiaxes $a$ and $a^{-1}$ where $a$ is very large. It looks like a long thin needle. The area is $\pi$, independent of $a$. But the integral $\int e(p)\,dp$ tends to $\infty$ as $a\to \infty$. Indeed, $e(p)\ge a$ for all $p$ (consider the distances from $0$ to the pointy vertices of ellipse), and the perimeter also grows, it's at least $4a$.