Parabola Halving a Triangle

The parabola and the line meet at the point with $x$ -coordinate $\xi$ , where $1 - \xi = a\xi^2$ , and hence $\xi \; = \; \frac{\sqrt{4a+1}-1}{2a}$ and we want $\begin{aligned} \int_0^\xi \big(1 - x - ax^2\big)\,dx & = \; \tfrac14 \\ \xi - \tfrac12\xi^2 - \tfrac13a\xi^3 & = \; \tfrac14 \\ 12a\big(\sqrt{4a+1}-1\big) - 3\big(\sqrt{4a+1}-1\big)^2 - \big(\sqrt{4a+1}-1\big)^3 & = \; 6a^2 \\ 2(4a+1)\sqrt{4a+1} & = \; 6a^2 + 12a + 2 \\ 9a^4 - 28a^3 - 6a^2 & = \; 0 \\ a^2\big[(9a - 14)^2 - 250\big] & = \; 0 \end{aligned}$ and hence, since $a$ is positive, $a \; = \; \frac{14 + 5\sqrt{10}}{9} \; = \; \boxed{3.31238}$

Denote $P(\beta,a\beta^2)$ as the point where the line and the parabola meet, such that $a\beta^2 = 1- \beta$

Clearly, the area of the triangle is $1/2$ and the parabola divides the are into two regions each with area $1/4$ .

$\begin{aligned} \int_0^\beta \big(1 - x - ax^2\big)\,dx & = \; \tfrac14 \\ \beta - \frac{\beta^2}{2} - \frac{a\beta^3}{3} & = \; \tfrac14 \\ \beta(12 - 6\beta - 4a\beta^2) & = \; 3 \\ \beta(12 - 6\beta - 4(1-\beta)) & = \; 3 \\ 2\beta^2 - 8\beta + 3 & = \; 0 \\ \beta = \frac{4-\sqrt{10}}{2} \end{aligned}$

Note that $\beta = \frac{4+\sqrt{10}}{2}$ was not considered since we want $0<=\beta<=1$ . Finally, $\begin{aligned} a = \frac{1-\beta}{\beta^2} & = \; \frac{14 + 5\sqrt{10}}{9} \; \approx \; \boxed{3.31238} \end{aligned}$