[THOUGHT] Let me analyze this problem step by step.

1. The ellipse has equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $0 < b < a$.
2. The parabola has equation $y^2 = 4px$ with $p > 0$.
3. The focus of the parabola is at $(p, 0)$.
4. One focus of the ellipse is at $(c, 0)$ where $c = \sqrt{a^2 - b^2}$.
5. These foci coincide, so $c = p$, meaning $\sqrt{a^2 - b^2} = p$.

The intersection points $P_1$ and $P_2$ of the parabola and ellipse are symmetric about the x-axis (since both curves are symmetric about the x-axis).

For a point $(x_0, y_0)$ on the parabola $y^2 = 4px$, the tangent line is: $yy_0 = 2p(x + x_0)$

If $P_1 = (x_0, y_0)$ and $P_2 = (x_0, -y_0)$ (by symmetry), then:

• Tangent at $P_1$: $yy_0 = 2p(x + x_0)$
• Tangent at $P_2$: $-yy_0 = 2p(x + x_0)$, or $yy_0 = -2p(x + x_0)$

To find intersection point $A$, we solve: $yy_0 = 2p(x + x_0)$ $-yy_0 = 2p(x + x_0)$

Adding these: $0 = 4p(x + x_0)$, so $x = -x_0$. Substituting back: $yy_0 = 2p(0) = 0$, so $y = 0$.

Thus $A = (-x_0, 0)$.

Now I need to find the angle $\angle P_1 A P_2 = 60°$.

$P_1 = (x_0, y_0)$, $P_2 = (x_0, -y_0)$, $A = (-x_0, 0)$.

Vector $\vec{AP_1} = (x_0 - (-x_0), y_0 - 0) = (2x_0, y_0)$ Vector $\vec{AP_2} = (2x_0, -y_0)$

The angle between them: $\cos(60°) = \frac{\vec{AP_1} \cdot \vec{AP_2}}{|\vec{AP_1}||\vec{AP_2}|}$

$\vec{AP_1} \cdot \vec{AP_2} = 4x_0^2 - y_0^2$ $|\vec{AP_1}| = |\vec{AP_2}| = \sqrt{4x_0^2 + y_0^2}$

$\frac{1}{2} = \frac{4x_0^2 - y_0^2}{4x_0^2 + y_0^2}$

This gives us: $4x_0^2 + y_0^2 = 2(4x_0^2 - y_0^2)$ $4x_0^2 + y_0^2 = 8x_0^2 - 2y_0^2$ $3y_0^2 = 4x_0^2$ $y_0^2 = \frac{4x_0^2}{3}$

Since $(x_0, y_0)$ is on the parabola: $y_0^2 = 4px_0$ So $\frac{4x_0^2}{3} = 4px_0$ $x_0 = 3p$

And $y_0^2 = 4p \cdot 3p = 12p^2$

Now $(x_0, y_0)$ is also on the ellipse: $\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1$ $\frac{9p^2}{a^2} + \frac{12p^2}{b^2} = 1$

Also, $b^2 = a^2 - p^2$ (since $c = p$).

Let me solve this system using Python. [/THOUGHT] [PYTHON] from sympy import symbols, solve, sqrt, Rational

a, p = symbols('a p', positive=True, real=True)

b^2 = a^2 - p^2

b_squared = a2 - p2

The point (x0, y0) = (3p, sqrt(12)*p) is on the ellipse

x0 = 3p y0_squared = 12p**2

Ellipse equation: x0^2/a^2 + y0^2/b^2 = 1

ellipse_eq = x02/a2 + y0_squared/b_squared - 1

Solve for a in terms of p

solutions = solve(ellipse_eq, a) print("Solutions for a:", solutions)

We need a > p (since b^2 = a^2 - p^2 > 0 means a > p)

and also a > b, which is a > sqrt(a^2 - p^2), always true for a > p

for sol in solutions: print(f"\nSolution: a = {sol}") # Calculate a^3/p^3 ratio = (sol/p)**3 ratio_simplified = ratio.simplify() print(f"a^3/p^3 = {ratio_simplified}")

# Numerical value
import math
# Let's substitute p = 1 to get numerical value
a_val = sol.subs(p, 1)
print(f"When p=1, a = {a_val}")
a_num = float(a_val)
print(f"Numerical a = {a_num}")
ratio_num = a_num**3
print(f"Numerical a^3/p^3 = {ratio_num}")
print(f"Floor of a^3/p^3 = {math.floor(ratio_num)}")

[/PYTHON]

[THOUGHT] From the Python output, we have two solutions for $a$:

1. $a = p(-2 + \sqrt{7})$ which gives a negative value for $a$ (since $\sqrt{7} \approx 2.646 < 2$ is false, actually $\sqrt{7} \approx 2.646 > 2$, so this is positive but less than $p$, which violates $a > p$ needed for $b^2 = a^2 - p^2 > 0$).
2. $a = p(2 + \sqrt{7})$ which gives $a \approx 4.646p$, satisfying $a > p$.

For the valid solution $a = p(2 + \sqrt{7})$:

• $\frac{a^3}{p^3} = (2 + \sqrt{7})^3 \approx 100.269$
• $\left\lfloor \frac{a^3}{p^3} \right\rfloor = 100$ [/THOUGHT]

\boxed{100}