Area under curve equals product of arc length and its projection

Question

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

Answer 1

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 $$f(x)=\cosh x+\frac{(b-a-1) (\sinh b-\sinh a)}{b-a}$$