[THOUGHT] We need to interpret the geometry and derive relationships.

Given:

• Ellipse: x^2/a^2 + y^2/b^2 = 1, with 0 < b < a.
• Parabola: y^2 = 4px, p > 0.
• One focus of ellipse coincides with focus of parabola.

For parabola y^2 = 4px, focus is at (p, 0).

For ellipse x^2/a^2 + y^2/b^2 = 1 (horizontal major axis), foci are at (±c, 0) where c^2 = a^2 - b^2.

Given one focus coincides: we can take c = p (positive focus). So p = c = sqrt(a^2 - b^2). So b^2 = a^2 - p^2.

Thus ellipse: x^2/a^2 + y^2/(a^2 - p^2) = 1.

Now find intersection points of parabola and ellipse.

Parabola: y^2 = 4px.

Substitute into ellipse: x^2/a^2 + (4px)/(a^2 - p^2) = 1.

Let’s denote D = a^2 - p^2 > 0.

Equation: x^2/a^2 + (4p x)/D = 1.

Multiply by a^2: x^2 + (4p a^2 / D) x = a^2.

So quadratic: x^2 + (4p a^2 / D) x - a^2 = 0.

Let’s compute D = a^2 - p^2.

We need intersection points P1, P2. They are symmetric about x-axis? Parabola y^2 = 4px, ellipse symmetric about x-axis, so intersections are symmetric: (x1, y1) and (x1, -y1) if x1 same. But quadratic in x gives possibly two x values? Wait, the equation is in x only, but for each x, y = ±√(4px). However, substituting y^2 gives equation in x only. That yields possible x coordinates. Since parabola is only defined for x ≥ 0, and ellipse symmetric, intersections could be two points with same x (if y ≠ 0) or possibly two distinct x values. Let's solve.

Let’s solve quadratic for x.

We have: x^2 + (4p a^2 / (a^2 - p^2)) x - a^2 = 0.

Let’s denote k = 4p a^2 / (a^2 - p^2).

Then x = [-k ± sqrt(k^2 + 4a^2)]/2.

But we need x ≥ 0 for parabola.

We'll compute later.

But note: The intersection points are P1 and P2. The tangent lines to the parabola at P1 and P2 intersect at A. Angle P1 A P2 = 60°.

We need to find relationship between a and p, then compute floor(a^3/p^3).

Let’s derive geometry.

Let intersection points: (x0, y0) and (x0, -y0) if same x? Actually, from quadratic, there could be two distinct x values. Let's check symmetry: The parabola y^2 = 4px is symmetric about x-axis. Ellipse is symmetric about x-axis and y-axis. Intersection points: If (x, y) is intersection, then (x, -y) also intersection. So for each x solution, we get two symmetric points unless y=0. So there could be two distinct x solutions, each giving a pair of symmetric points, total four intersections. But the problem says "the two intersection points of the parabola and the ellipse be denoted by P1 and P2". That suggests only two intersection points total. Possibly they mean the two distinct points of intersection (maybe symmetric about x-axis? Actually if two intersection points total, then they must be symmetric about x-axis and have same x coordinate, so y = ±y0. That would mean the quadratic in x has a single positive root x0 (double root) or perhaps only one root? But quadratic generally has two roots. Let's examine.

Set equation: x^2/a^2 + (4p x)/(a^2 - p^2) = 1.

Multiply by a^2: x^2 + (4p a^2/(a^2 - p^2)) x - a^2 = 0.

Discriminant: Δ = (4p a^2/(a^2 - p^2))^2 + 4a^2.

This is > 0, so two distinct real roots. One root might be negative, one positive. Since parabola x ≥ 0, only positive root(s) correspond to intersections. Possibly only one positive root, and for that x, y = ±√(4px) gives two points symmetric about x-axis. That would give two intersection points P1 and P2 symmetric about x-axis. That matches "two intersection points" total. So likely the quadratic has one positive root and one negative root. Then P1 = (x0, y0), P2 = (x0, -y0).

Let's verify: For ellipse and parabola intersecting, typically there can be up to 4 intersections. But maybe given b < a and focus coincidence, only two intersections occur.

We'll assume that.

Thus P1 = (x0, y0), P2 = (x0, -y0) with y0 = √(4p x0) > 0.

Now tangent lines to parabola at these points.

Parabola y^2 = 4px. Differentiate implicitly: 2y dy/dx = 4p => dy/dx = 2p / y.

At (x0, y0): slope m1 = 2p / y0.

At (x0, -y0): slope m2 = 2p / (-y0) = -2p / y0.

Equation of tangent at P1: y - y0 = m1 (x - x0).

Similarly at P2: y + y0 = m2 (x - x0).

Find intersection point A of these two lines.

Solve:

Line1: y = y0 + m1 (x - x0) = y0 + (2p/y0)(x - x0).

Line2: y = -y0 + m2 (x - x0) = -y0 + (-2p/y0)(x - x0) = -y0 - (2p/y0)(x - x0).

Set equal:

y0 + (2p/y0)(x - x0) = -y0 - (2p/y0)(x - x0).

=> y0 + (2p/y0)(x - x0) + y0 + (2p/y0)(x - x0) = 0.

=> 2y0 + (4p/y0)(x - x0) = 0.

=> (4p/y0)(x - x0) = -2y0.

=> x - x0 = -2y0 * (y0/(4p)) = - (2y0^2)/(4p) = - y0^2/(2p).

But y0^2 = 4p x0.

So x - x0 = - (4p x0)/(2p) = -2x0.

Thus x_A = x0 - 2x0 = -x0.

So A = (-x0, y_A). Find y_A from line1:

y_A = y0 + (2p/y0)(-x0 - x0) = y0 + (2p/y0)(-2x0) = y0 - (4p x0)/y0.

But (4p x0)/y0 = y0^2 / y0 = y0.

So y_A = y0 - y0 = 0.

Thus A = (-x0, 0).

So A lies on x-axis at x = -x0.

Now we have triangle P1 A P2: vertices P1 = (x0, y0), P2 = (x0, -y0), A = (-x0, 0).

This triangle is isosceles with A at apex? Actually P1 and P2 symmetric about x-axis, A on x-axis. So base P1P2 is vertical segment at x0 from y0 to -y0, length 2y0. Vertex A is at (-x0, 0). So sides AP1 and AP2 are equal.

Angle P1 A P2 is the angle at A between lines AP1 and AP2.

Compute vectors:

AP1 = (x0 - (-x0), y0 - 0) = (2x0, y0).

AP2 = (x0 - (-x0), -y0 - 0) = (2x0, -y0).

Angle between these vectors: use dot product.

AP1 · AP2 = (2x0)(2x0) + y0(-y0) = 4x0^2 - y0^2.

|AP1| = sqrt((2x0)^2 + y0^2) = sqrt(4x0^2 + y0^2).

Similarly |AP2| = same.

So cos(angle) = (4x0^2 - y0^2) / (4x0^2 + y0^2).

Given angle = 60°, cos60° = 1/2.

Thus (4x0^2 - y0^2) / (4x0^2 + y0^2) = 1/2.

Cross multiply: 2(4x0^2 - y0^2) = 4x0^2 + y0^2.

=> 8x0^2 - 2y0^2 = 4x0^2 + y0^2.

=> 8x0^2 - 4x0^2 = y0^2 + 2y0^2.

=> 4x0^2 = 3y0^2.

=> y0^2 = (4/3) x0^2.

But y0^2 = 4p x0.

Thus 4p x0 = (4/3) x0^2.

If x0 ≠ 0 (since intersection points distinct, x0 > 0), divide both sides by 4x0: p = (1/3) x0.

Thus x0 = 3p.

Now we need to ensure x0 is the positive root of quadratic earlier.

Recall quadratic: x^2 + (4p a^2/(a^2 - p^2)) x - a^2 = 0.

Plug x0 = 3p.

Compute coefficients.

Let D = a^2 - p^2.

Equation: x^2 + (4p a^2 / D) x - a^2 = 0.

Plug x = 3p:

(3p)^2 + (4p a^2 / D)*(3p) - a^2 = 0.

=> 9p^2 + (12 p^2 a^2 / D) - a^2 = 0.

Multiply by D: 9p^2 D + 12 p^2 a^2 - a^2 D = 0.

But D = a^2 - p^2.

So 9p^2 (a^2 - p^2) + 12 p^2 a^2 - a^2 (a^2 - p^2) = 0.

Compute: 9p^2 a^2 - 9p^4 + 12p^2 a^2 - a^4 + p^2 a^2 = 0.

Combine p^2 a^2 terms: 9p^2 a^2 + 12p^2 a^2 + p^2 a^2 = 22 p^2 a^2.

So: 22 p^2 a^2 - 9p^4 - a^4 = 0.

Rearrange: -a^4 + 22 p^2 a^2 - 9p^4 = 0.

Multiply by -1: a^4 - 22 p^2 a^2 + 9p^4 = 0.

Let u = a^2. Then u^2 - 22 p^2 u + 9p^4 = 0.

Solve: u = [22 p^2 ± sqrt(484 p^4 - 36 p^4)]/2 = [22 p^2 ± sqrt(448 p^4)]/2 = [22 p^2 ± p^2 sqrt(448)]/2.

sqrt(448) = sqrt(64*7) = 8√7.

Thus u = p^2 (22 ± 8√7)/2 = p^2 (11 ± 4√7).

So a^2 = p^2 (11 ± 4√7).

Since a > p (because b^2 = a^2 - p^2 > 0, so a^2 > p^2). Both 11 ± 4√7 are > p^2? Let's check: 4√7 ≈ 4*2.64575 = 10.583. So 11 + 10.583 = 21.583, and 11 - 10.583 = 0.417. So a^2 = p^2 * 21.583 or a^2 = p^2 * 0.417. Since a > p, a^2 > p^2, so we take a^2 = p^2 (11 + 4√7). The other root gives a^2 < p^2, which would make b^2 = a^2 - p^2 negative (since a^2 < p^2). But b^2 must be positive, so we need a^2 > p^2. Thus a^2 = p^2 (11 + 4√7).

Thus a^3 / p^3 = (a^2/p^2)^(3/2) = (11 + 4√7)^(3/2).

We need floor of that.

Compute numerically.

Let’s compute with Python. </think>

[PYTHON] import math

Compute a^2/p^2 = 11 + 4*sqrt(7)

sqrt7 = math.sqrt(7) ratio_sq = 11 + 4 * sqrt7 print("a^2 / p^2 =", ratio_sq)

Compute a/p = sqrt(ratio_sq)

ratio = math.sqrt(ratio_sq) print("a / p =", ratio)

Compute (a/p)^3

ratio_cubed = ratio ** 3 print("(a/p)^3 =", ratio_cubed)

Compute floor

floor_val = math.floor(ratio_cubed) print("floor =", floor_val)

Let's also compute with high precision to verify

import decimal decimal.getcontext().prec = 50 sqrt7_dec = decimal.Decimal(7).sqrt() ratio_sq_dec = decimal.Decimal(11) + decimal.Decimal(4) * sqrt7_dec ratio_dec = ratio_sq_dec.sqrt() ratio_cubed_dec = ratio_dec ** 3 print("Using decimal:") print("a^2/p^2 =", ratio_sq_dec) print("a/p =", ratio_dec) print("(a/p)^3 =", ratio_cubed_dec) floor_dec = int(ratio_cubed_dec) print("floor =", floor_dec) [/PYTHON]

[THOUGHT] From the Python output, we have (\left\lfloor \frac{a^3}{p^3} \right\rfloor = 100).
Thus the answer is 100.

[/THOUGHT] \boxed{100}