How to explain or give reasons about the curve?

Question

I know that arc length of the curve is: $$s=\int_{a}^{b}\sqrt{1+\big(f^{'}(x)\big)^2}dx$$ The question is "Is there a smooth curve y=f(x)" where length over the interval $0\leq x \leq a$, (where a is any constant) is always $\sqrt{2}a$.

How can we say about the curve $y=f(x)$ by putting $$\int_{0}^{a}\sqrt{1+\big(f^{'}(x)\big)^2}dx=\sqrt{2}a$$

Answer 1

Yes: take $y=x$

Trivially the integral is $\sqrt{2}a$ as required.

Answer 2

If you differentiate both members on $a$,

$$\sqrt{1+f'^2(a)}=\sqrt2,$$ which implies

$$f'(a)=\pm1$$ or $$f(a)=c\pm a.$$

These are line segments slanted by $45°$.