Area under curve equals product of arc length and its projection Find equation of a curve whose  area under it equals product of arc length  $L$ and its projection $(b-a)$ on the x-axis.
$$ A = L (b-a)$$
I was trying to establish a corollary to the Amazing Catenary property

 A: Catenary satisfies the given problem
Consider $f(x)=\cosh x+k$. We have
$$A=\int_a^b (k+\cosh x) \, dx=k (b-a)+\sinh b-\sinh a$$
and
$$(b-a)L=(b-a) \int_a^b \sqrt{1+\sinh ^2 x} \, dx=(b-a) \int_a^b \cosh x \, dx=(b-a) (\sinh b-\sinh a)$$
therefore for $k= \frac{(b-a-1) (\sinh b-\sinh a)}{b-a}$ we have $A=(b-a)L$.
we get the result $$f(x)=\cosh x+\frac{(b-a-1) (\sinh b-\sinh a)}{b-a}$$
A: Shift the function such that $ a \to 0$
$$ \int_0^x f(t) dt =x  \int_0^x \sqrt{ 1 + (f'(t))^2} dt $$
d.w.r.t.x
$$ f(x) =  \sqrt{ 1 + (f'(x))^2} + \frac{xf''(x)}{\sqrt{ 1 + f'(x)^2} }$$
Sub: $ f(x) \to y$
$$  y = \sqrt{1 + (y')^2} + \frac{ x y''}{ \sqrt{ 1 + y'^2} }$$
Now all you have to do is solve this DE.
A: A constant map satisfies the equation. Let's prove that there is no other solution.
Suppose that $f$ is continuously differentiable. As mentioned, if $f$ is a solution (with adequate translation) it satisfies the functional equation:
$$ \int_0^x f(t) dt =x  \int_0^x \sqrt{ 1 + (f^ \prime(t))^2} dt \tag{1}.$$
The RHS, namely
$$R(x)= x  \int_0^x \sqrt{ 1 + (f^ \prime(t))^2} dt$$ remains invariant if we change $f$ into $g(x) = f(0)  - \int_0^x f^\prime(t) \ dt$ as $g^\prime(x) = - f^\prime(x)$.
However if for a point $x_0$, we have $f^ \prime(x_0) \neq 0$, the LHS of the equation $(1)$ will change locally around $x_0$ if $f$ is replaced with $g$. A contradiction. As this is independent of $x_0$, we get $f^\prime(x) = 0$ for all $x \in \mathbb R$ meaning that $f$ has to be constant.
