Universal parabolic constant

Take ${\displaystyle y={\frac {x^{2}}{4f}}}$ as the equation of the parabola. The focal parameter is ${\displaystyle p=2f}$ and the semilatus rectum is ${\displaystyle \ell =2f}$.

${\displaystyle {\begin{aligned}P&:={\frac {1}{p}}\int _{-\ell }^{\ell }{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}\,dx\\&={\frac {1}{2f}}\int _{-2f}^{2f}{\sqrt {1+{\frac {x^{2}}{4f^{2}}}}}\,dx\\&=\int _{-1}^{1}{\sqrt {1+t^{2}}}\,dt\quad (x=2ft)\\&=\operatorname {arcsinh} (1)+{\sqrt {2}}\\&=\ln(1+{\sqrt {2}})+{\sqrt {2}}.\end{aligned}}}$

P is a transcendental number.

Proof. Suppose that P is algebraic. Then ${\displaystyle \!\ P-{\sqrt {2}}=\ln(1+{\sqrt {2}})}$ must also be algebraic. However, by the Lindemann–Weierstrass theorem, ${\displaystyle \!\ e^{\ln(1+{\sqrt {2}})}=1+{\sqrt {2}}}$ would be transcendental, which is not the case. Hence P is transcendental.

Since P is transcendental, it is also irrational.

The average distance from a point randomly selected in the unit square to its center is[5]

${\displaystyle d_{\text{avg}}={P \over 6}.}$

Proof.

${\displaystyle {\begin{aligned}d_{\text{avg}}&:=8\int _{0}^{1 \over 2}\int _{0}^{x}{\sqrt {x^{2}+y^{2}}}\,dy\,dx\\&=8\int _{0}^{1 \over 2}{1 \over 2}x^{2}(\ln(1+{\sqrt {2}})+{\sqrt {2}})\,dx\\&=4P\int _{0}^{1 \over 2}x^{2}\,dx\\&={P \over 6}.\end{aligned}}}$