Problem Analysis #23

[THOUGHT] The parabola is $y^2 = 4px$ with focus at $(p, 0)$. The ellipse has foci at $(\pm c, 0)$ where $c = \sqrt{a^2 - b^2}$, and one focus coincides with $(p, 0)$, so $c = p$ and $b^2 = a^2 - p^2$.

The intersection points satisfy $\frac{x^2}{a^2} + \frac{4px}{b^2} = 1$, a quadratic in $x$ with one positive root $x_0 > 0$. The points are $P_1(x_0, \sqrt{4px_0})$ and $P_2(x_0, -\sqrt{4px_0})$, corresponding to parameters $t$ and $-t$ on the parabola, with $x_0 = p t^2$.

The tangents at $t$ and $-t$ intersect at $A = (-p t^2, 0)$.

Vectors $\overrightarrow{AP_1} = 2pt (t, 1)$, $\overrightarrow{AP_2} = 2pt (t, -1)$. The angle $\phi = \angle P_1 A P_2$ satisfies $\cos \phi = \frac{t^2 - 1}{t^2 + 1}$. Given $\phi = 60^\circ$, $\cos 60^\circ = 0.5$, so $\frac{t^2 - 1}{t^2 + 1} = \frac{1}{2}$, yielding $t^2 = 3$, $t = \sqrt{3}$ (since $t > 0$).

Substitute into ellipse: $\frac{(p \cdot 3)^2}{a^2} + \frac{(2p\sqrt{3})^2}{b^2} = 1$, so $\frac{9p^2}{a^2} + \frac{12p^2}{b^2} = 1$.

Let $r = a^2 / p^2 > 1$, then $b^2 / p^2 = r - 1$. The equation becomes $\frac{9}{r} + \frac{12}{r-1} = 1$. Solving: $9(r-1) + 12r = r(r-1)$, $21r - 9 = r^2 - r$, $r^2 - 22r + 9 = 0$, $r = 11 + 4\sqrt{7}$ (positive root).

Then $\frac{a^3}{p^3} = r^{3/2}$. Compute numerically to find floor. [/THOUGHT] [PYTHON] import math

sqrt7 = math.sqrt(7) r = 11 + 4 * sqrt7 val = r ** 1.5 print(val) print(math.floor(val)) [/PYTHON]

[THOUGHT] The Python output gives r^{3/2} ≈ 100.269, so floor is 100. [/THOUGHT] $\boxed{100}$

Explanation

The reasoning follows a logical sequence to derive the value of (\left\lfloor \frac{a^3