[THOUGHT] We have an ellipse: x^2/a^2 + y^2/b^2 = 1 with 0 < b < a. One focus of the ellipse coincides with the focus of the parabola y^2 = 4px (p > 0). The parabola's standard form: y^2 = 4px has focus at (p, 0) and vertex at (0,0), axis along x-axis. So the ellipse's focus that coincides with parabola's focus is at (p, 0). For an ellipse x^2/a^2 + y^2/b^2 = 1 with a > b > 0, the foci are at (±c, 0) where c^2 = a^2 - b^2. Since the focus is at (p, 0) and p > 0, we have c = p. So p^2 = a^2 - b^2. (1)

The ellipse and parabola intersect at two points P1 and P2. Let A be intersection of tangents to parabola at P1 and P2. Given ∠P1 A P2 = 60°.

We need floor(a^3 / p^3).

We need to relate the geometry. Let's parameterize parabola: For parabola y^2 = 4px, we can write parametric form: (pt^2, 2pt) where t is parameter. Then point P = (pt^2, 2pt). The tangent line to parabola at (pt^2, 2pt) has equation: y * (2pt) = 2p(x + pt^2) ??? Actually standard tangent to parabola y^2 = 4ax at (at^2, 2at) is ty = x + at^2. Here a = p, so tangent at (pt^2, 2pt) is t y = x + p t^2.

Let P1 correspond to parameter t1, P2 correspond to parameter t2. Then tangents at P1 and P2: t1 y = x + p t1^2 t2 y = x + p t2^2

Intersection point A of these two lines: solve for (x,y). Subtract: (t1 - t2) y = p(t1^2 - t2^2) = p(t1 - t2)(t1 + t2). If t1 ≠ t2, then y = p(t1 + t2). Then from first equation: t1 * p(t1 + t2) = x + p t1^2 => x = p t1(t1 + t2) - p t1^2 = p t1 t2.

So A = (p t1 t2, p(t1 + t2)).

Now P1 = (p t1^2, 2p t1), P2 = (p t2^2, 2p t2).

Given ∠P1 A P2 = 60°. That's angle at A between vectors A->P1 and A->P2.

Compute vectors: v1 = P1 - A = (p t1^2 - p t1 t2, 2p t1 - p(t1 + t2)) = (p t1(t1 - t2), p(2t1 - t1 - t2)) = (p t1(t1 - t2), p(t1 - t2)). Similarly v2 = P2 - A = (p t2^2 - p t1 t2, 2p t2 - p(t1 + t2)) = (p t2(t2 - t1), p(2t2 - t1 - t2)) = (p t2(t2 - t1), p(t2 - t1)).

Simplify: v1 = p(t1 - t2) (t1, 1). Actually careful: v1 = (p t1(t1 - t2), p(t1 - t2)) = p(t1 - t2) (t1, 1). v2 = (p t2(t2 - t1), p(t2 - t1)) = p(t2 - t1) (t2, 1) = - p(t1 - t2) (t2, 1).

So v1 = k (t1, 1) where k = p(t1 - t2). v2 = -k (t2, 1).

Angle between v1 and v2: cos θ = (v1·v2) / (|v1| |v2|). Since θ = 60°, cos 60° = 1/2.

Compute dot product: v1·v2 = k (t1,1) · (-k (t2,1)) = -k^2 (t1 t2 + 1). |v1| = |k| sqrt(t1^2 + 1). |v2| = |k| sqrt(t2^2 + 1). So |v1| |v2| = k^2 sqrt((t1^2+1)(t2^2+1)).

Thus cos θ = - (t1 t2 + 1) / sqrt((t1^2+1)(t2^2+1)) = 1/2.

So we have:

• (t1 t2 + 1) / sqrt((t1^2+1)(t2^2+1)) = 1/2. Multiply both sides by -1: (t1 t2 + 1) / sqrt((t1^2+1)(t2^2+1)) = -1/2.

Thus t1 t2 + 1 = - (1/2) sqrt((t1^2+1)(t2^2+1)).

Square both sides: (t1 t2 + 1)^2 = (1/4) (t1^2+1)(t2^2+1).

Expand: t1^2 t2^2 + 2 t1 t2 + 1 = (1/4)(t1^2 t2^2 + t1^2 + t2^2 + 1).

Multiply by 4: 4 t1^2 t2^2 + 8 t1 t2 + 4 = t1^2 t2^2 + t1^2 + t2^2 + 1.

Bring all: 4 t1^2 t2^2 - t1^2 t2^2 + 8 t1 t2 + 4 - 1 - t1^2 - t2^2 = 0 => 3 t1^2 t2^2 + 8 t1 t2 + 3 - t1^2 - t2^2 = 0.

Rearrange: 3 t1^2 t2^2 - t1^2 - t2^2 + 8 t1 t2 + 3 = 0.

Group terms: (3 t1^2 t2^2 - t1^2 - t2^2) + 8 t1 t2 + 3 = 0.

Factor first part: t1^2 (3 t2^2 - 1) - t2^2 + 8 t1 t2 + 3 = 0. Not obvious.

Alternatively, we can treat as quadratic in t1? But maybe we can also use that P1 and P2 lie on ellipse.

Since P1 and P2 are intersection points of parabola and ellipse, they satisfy ellipse equation: (x^2 / a^2) + (y^2 / b^2) = 1. For point (p t^2, 2p t): (p^2 t^4 / a^2) + (4 p^2 t^2 / b^2) = 1. Divide by p^2: (t^4 / (a^2/p^2)) + (4 t^2 / (b^2/p^2)) = 1/p^2. But we have p^2 = a^2 - b^2. Let’s denote e = c/a = p/a (since c = p). Then p = a e, and b^2 = a^2 - p^2 = a^2(1 - e^2). So b = a sqrt(1 - e^2). Then p = a e.

We want a^3 / p^3 = (a/p)^3 = (1/e)^3.

Thus we need to find e (eccentricity of ellipse) from conditions.

From ellipse focus: p = a e, so e = p/a. Since p = c, e = c/a = p/a.

Now substitute into ellipse equation for parameter t: (p^2 t^4)/a^2 + (4 p^2 t^2)/b^2 = 1. But p^2 = a^2 e^2, b^2 = a^2 (1 - e^2). So: (a^2 e^2 t^4)/a^2 + (4 a^2 e^2 t^2)/(a^2 (1 - e^2)) = 1 => e^2 t^4 + (4 e^2 t^2)/(1 - e^2) = 1. Multiply by (1 - e^2): e^2 t^4 (1 - e^2) + 4 e^2 t^2 = 1 - e^2. Rearrange: e^2 (1 - e^2) t^4 + 4 e^2 t^2 - (1 - e^2) = 0. This is a quadratic in t^2: Let u = t^2. Then e^2 (1 - e^2) u^2 + 4 e^2 u - (1 - e^2) = 0. Thus (e^2 (1 - e^2)) u^2 + 4 e^2 u - (1 - e^2) = 0.

The two intersection points correspond to two distinct real t values, so two distinct u values (maybe positive? Actually t can be positive or negative, but u = t^2 >=0). For two intersection points, there are two distinct t values, so the quadratic in u should have two positive roots? Actually if t1 and t2 are distinct, then u1 = t1^2 and u2 = t2^2 are positive unless t is zero. But they could be same if t1 = -t2? That would give same u but different t. But then intersection points symmetric about x-axis. However we have two intersection points P1 and P2, they could be symmetric. Let's see.

From the quadratic: A u^2 + B u + C = 0, where A = e^2 (1 - e^2), B = 4 e^2, C = -(1 - e^2). Since A > 0? e in (0,1) because ellipse with a > b > 0, e = c/a = p/a, p>0, a>p? Actually p = a e, and since p < a? Because c = p < a? For ellipse, c < a, so e < 1. Also e > 0. So 0 < e < 1. Then 1 - e^2 > 0, so A > 0, C < 0. So product of roots = C/A = - (1 - e^2) / (e^2 (1 - e^2)) = -1/e^2 < 0. So one root positive, one negative. That means u1 and u2 have opposite signs. But u = t^2 >=0, so only non-negative roots allowed. So there is only one positive root? That would give two t values: t = ± sqrt(u). So intersection points are symmetric about x-axis: P1 and P2 are symmetric: t2 = -t1. Then t1 t2 = - t1^2, and t1 + t2 = 0. That simplifies things.

Let's check: if t2 = -t1, then from earlier equation (t1 t2 + 1) / sqrt((t1^2+1)(t2^2+1)) = -1/2. With t2 = -t1, t1 t2 = -t1^2, and t2^2 = t1^2. So LHS = (-t1^2 + 1) / sqrt((t1^2+1)(t1^2+1)) = (1 - t1^2) / (t1^2+1). Set equal to -1/2: (1 - t1^2)/(t1^2+1) = -1/2 => Cross multiply: 2(1 - t1^2) = - (t1^2+1) => 2 - 2 t1^2 = - t1^2 - 1 => 2 + 1 = 2 t1^2 - t1^2 => 3 = t1^2 => t1^2 = 3. So t1 = sqrt(3) or -sqrt(3). Then t2 = -t1. So indeed symmetric.

Thus we have t1^2 = 3. Then u = t1^2 = 3 is a root of the quadratic in u. Substitute u=3 into quadratic to get equation for e.

Quadratic: e^2 (1 - e^2) u^2 + 4 e^2 u - (1 - e^2) = 0. Plug u=3: e^2 (1 - e^2) * 9 + 4 e^2 * 3 - (1 - e^2) = 0 => 9 e^2 (1 - e^2) + 12 e^2 - (1 - e^2) = 0. Expand: 9 e^2 - 9 e^4 + 12 e^2 - 1 + e^2 = 0 => Combine e^2 terms: 9 e^2 + 12 e^2 + e^2 = 22 e^2. So: 22 e^2 - 9 e^4 - 1 = 0 => Multiply by -1: 9 e^4 - 22 e^2 + 1 = 0. Solve for e^2: Let x = e^2. Then 9x^2 - 22x + 1 = 0. Discriminant: 22^2 - 491 = 484 - 36 = 448 = 647? Actually 448 = 64 * 7? 647=448 yes. So sqrt(448) = sqrt(647)=8 sqrt(7). Thus x = (22 ± 8 sqrt(7)) / (29) = (22 ± 8 sqrt(7))/18 = (11 ± 4 sqrt(7))/9.

We have e^2 = (11 ± 4 sqrt(7))/9. Since e < 1, e^2 < 1. Check both possibilities: (11 + 4 sqrt(7))/9 ≈ (11 + 4*2.6458)/9 = (11 + 10.5832)/9 = 21.5832/9 ≈ 2.398 >1, so discard. Thus e^2 = (11 - 4 sqrt(7))/9 ≈ (11 - 10.5832)/9 = 0.4168/9 ≈ 0.04631? Wait compute: 11 - 10.5832 = 0.4168, divided by 9 = 0.04631. That seems too small. Let's compute more precisely: sqrt(7) ≈ 2.6457513110645907. Then 4 sqrt(7) ≈ 10.583005244258363. So 11 - 10.583005244258363 = 0.416994755741637. Divide by 9: 0.04633275063795967. So e^2 ≈ 0.04633, e ≈ 0.2152. That is less than 1, plausible.

Thus e = sqrt((11 - 4 sqrt(7))/9). Then a/p = 1/e = 1 / sqrt((11 - 4 sqrt(7))/9) = sqrt(9/(11 - 4 sqrt(7))) = 3 / sqrt(11 - 4 sqrt(7)).

We need (a/p)^3 = (1/e)^3 = (3 / sqrt(11 - 4 sqrt(7)))^3 = 27 / (11 - 4 sqrt(7))^(3/2). But we can compute numerically.

Alternatively, rationalize: sqrt(11 - 4 sqrt(7)). Maybe we can simplify: Suppose sqrt(11 - 4 sqrt(7)) = sqrt(a) - sqrt(b) with a,b rational? (sqrt(a)-sqrt(b))^2 = a+b -2 sqrt(ab) = 11 - 4 sqrt(7). So a+b=11, 2 sqrt(ab)=4 sqrt(7) => sqrt(ab)=2 sqrt(7) => ab=28. So a,b are roots of t^2 -11t+28=0 => (t-4)(t-7)=0 => a=7,b=4 or a=4,b=7. So sqrt(11 - 4 sqrt(7)) = sqrt(7) - sqrt(4) = sqrt(7) - 2 (since sqrt(7) > 2). Check: (sqrt(7)-2)^2 = 7+4 -4 sqrt(7)=11-4 sqrt(7). Yes.

Thus sqrt(11 - 4 sqrt(7)) = sqrt(7) - 2.

Then a/p = 3 / (sqrt(7) - 2) = 3 (sqrt(7)+2) / (7-4) = 3 (sqrt(7)+2)/3 = sqrt(7) + 2.

So a/p = sqrt(7) + 2.

Thus (a/p)^3 = (sqrt(7)+2)^3.

Compute: (sqrt(7)+2)^3 = (sqrt(7)+2)^2 * (sqrt(7)+2) = (7 + 4 sqrt(7) + 4) * (sqrt(7)+2) = (11 + 4 sqrt(7)) (sqrt(7)+2) = 11 sqrt(7) + 22 + 4*7 + 8 sqrt(7) = 11 sqrt(7) + 8 sqrt(7) + 22 + 28 = 19 sqrt(7) + 50.

So a^3 / p^3 = 50 + 19 sqrt(7). Approximately sqrt(7) ≈ 2.6457513110645907, so 19 sqrt(7) ≈ 50.269274901, plus 50 = 100.269274901. So floor is 100.

Thus answer should be 100.

We should verify with Python to ensure no mistake.

We'll compute e from equation, compute a/p, cube it, and take floor.

[/THOUGHT] [PYTHON] import sympy as sp

Define variable e (eccentricity)

e = sp.symbols('e', positive=True)

Equation from quadratic root u=3: 9 e^2 (1 - e^2) + 12 e^2 - (1 - e^2) = 0

equation = 9e**2(1 - e2) + 12*e2 - (1 - e**2) solution = sp.solve(equation, e) print("Solutions for e:", solution)

Filter e in (0,1)

valid_e = [sol for sol in solution if 0 < sol < 1] print("Valid e:", valid_e) e_val = valid_e[0] print("e =", e_val)

a/p = 1/e

a_over_p = 1/e_val print("a/p =", a_over_p)

Simplify expression

a_over_p_simplified = sp.simplify(a_over_p) print("Simplified a/p =", a_over_p_simplified)

Compute cube

cube = a_over_p_simplified**3 print("(a/p)^3 =", cube)

Simplify cube

cube_simplified = sp.simplify(cube) print("Simplified (a/p)^3 =", cube_simplified)

Evaluate numerically

cube_num = cube_simplified.evalf() print("Numerical value:", cube_num)

Floor

floor_val = int(sp.floor(cube_num)) print("Floor:", floor_val)

Also compute directly from derived expression: sqrt(7)+2

sqrt7 = sp.sqrt(7) derived = (sqrt7 + 2)**3 print("Derived (sqrt(7)+2)^3 =", derived) print("Simplified derived:", sp.simplify(derived)) print("Numerical derived:", derived.evalf()) print("Floor derived:", int(sp.floor(derived))) [/PYTHON]

[THOUGHT] The Python output confirms that (a/p = 2 + \sqrt{7}), so ((a/p)^3 = (2 + \sqrt{7})^3 \approx 100.269). The floor is 100. [/THOUGHT] \boxed{100}