Let A₁ be the area of the parabola y²=4ax lying between vertex and latus rectum and A₂ be the area between latus rectum and double ordinate x=2a. Then A₁/A₂=

A₁ be the area of the parabola y²=4ax lying between vertex and latus rectum

${\mathrm{A}}_1=2{\int }_0^{\mathrm{a}} \sqrt{4\mathrm{ax}}dx=\frac{8{\mathrm{a}}^2}{3}$

A₂ be the area between latus rectum and double ordinate x=2a.

$A_2=2\left[{\int }_0^{2a} \sqrt{4ax}-{\int }_0^a \sqrt{4ax}\right]=\frac{16\sqrt{2}}{2}{a}^2-\frac{8{\mathrm{a}}^2}{3}==\frac{8{\mathrm{a}}^2}{3}(2\sqrt{2}-1)$

$\Rightarrow \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{1}{2\sqrt{2}-1}=\frac{2\sqrt{2}+1}{7}$