[PYTHON] import math

Calculate 50 + 19 * sqrt(7)

val = 50 + 19 * math.sqrt(7) print(f"{val=}") print(f"{math.floor(val)=}") [/PYTHON] ps

1. Identify the geometric setup:

   • Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($0 < b < a$).
   • Parabola: $y^2 = 4px$ ($p > 0$).
   • Focus of parabola: $F_p = (p, 0)$.
   • Foci of ellipse: $F_{e} = (\pm c, 0)$ where $c = \sqrt{a^2 - b^2}$.
   • Condition: One focus coincides. Since $p>0$, it must be the positive focus $(c, 0)$. So $c = p$.
   • Thus $p^2 = a^2 - b^2 \implies b^2 = a^2 - p^2$.

2. Intersection points $P_1, P_2$:

   • Substitute $y^2 = 4px$ into ellipse equation: $\frac{x^2}{a^2} + \frac{4px}{a^2 - p^2} = 1$.
   • Let $u = a/p$. Then $a = up$.
   • $\frac{x^2}{u^2 p^2} + \frac{4px}{p^2(u^2 - 1)} = 1$.
   • $\frac{x^2}{u^2 p^2} + \frac{4x}{p(u^2 - 1)} - 1 = 0$.
   • Multiply by $u^2 p^2 (u^2 - 1)$: $(u^2 - 1) x^2 + 4 p u^2 x - u^2 p^2 (u^2 - 1) = 0$.
   • This gives the x-coordinates of intersection. Let the positive root be $x_0$.

3. Tangent intersection $A$:

   • Tangents at $P_1(x_0, y_0)$ and $P_2(x_0, -y_0)$ to the parabola $y^2 = 4px$.
   • Tangent at $(x_0, y_0)$: $y y_0 = 2p(x + x_0)$.
   • Tangent at $(x_0, -y_0)$: $-y y_0 = 2p(x + x_0)$.
   • Intersection $A$ is at $x = -x_0, y = 0$. So $A = (-x_0, 0)$.

4. Angle condition:

   • $\angle P_1 A P_2 = 60^\circ$.
   • Since $A$ is on x-axis and $P_1, P_2$ are symmetric, $\angle P_1 A X = 30^\circ$.
   • Slope of $A P_1$ is $\tan 30^\circ = 1/\sqrt{3}$.
   • Slope is $\frac{y_0 - 0}{x_0 - (-x_0)} = \frac{y_0}{2x_0}$.
   • So $\frac{y_0}{2x_0} = \frac{1}{\sqrt{3}} \implies 3 y_0^2 = 4 x_0^2$.
   • Since $P_1$ is on parabola, $y_0^2 = 4px_0$.
   • $3(4px_0) = 4x_0^2 \implies 3p = x_0$ (since $x_0 \neq 0$).

5. Solve for $a/p$:

   • Substitute $x_0 = 3p$ into the intersection equation: $(u^2 - 1) (3p)^2 + 4 p u^2 (3p) - u^2 p^2 (u^2 - 1) = 0$.
   • Divide by $p^2$: $9(u^2 - 1) + 12 u^2 - u^2 (u^2 - 1) = 0$. $9u^2 - 9 + 12u^2 - u^4 + u^2 = 0$. $-u^4 + 22u^2 - 9 = 0$. $u^4 - 22u^2 + 9 = 0$.
   • Solve for $u^2$: $u^2 = \frac{22 \pm \sqrt{484 - 36}}{2} = 11 \pm 4\sqrt{7}$.
   • Since $b^2 = a^2 - p^2 = p^2(u^2 - 1) > 0$, we need $u^2 > 1$.
   • $11 - 4\sqrt{7} \approx 11 - 10.58 = 0.42 < 1$.
   • $11 + 4\sqrt{7} \approx 21.58 > 1$.
   • So $u^2 = 11 + 4\sqrt{7}$.

6. Calculate target value:

   • We need $\lfloor u^3 \rfloor$.
   • $u = \sqrt{11 + 4\sqrt{7}} = \sqrt{(\sqrt{7} + 2)^2} = 2 + \sqrt{7}$.
   • $u^3 = (2 + \sqrt{7})^3 = 8 + 12\sqrt{7} + 42 + 7\sqrt{7} = 50 + 19\sqrt{7}$.
   • $19\sqrt{7} \approx 19 \times 2.64575 = 50.269$.
   • $u^3 \approx 100.269$.
   • Floor is 100.

The Python code confirmed the calculation. $val = 100.269...$ $\lfloor val \rfloor = 100$.

Double check if there are any other constraints. $0 < b < a$. $b^2 = a^2 - p^2 > 0 \implies a > p \implies u > 1$. Checked. $b < a \implies \sqrt{a^2 - p^2} < a \implies a^2 - p^2 < a^2 \implies -p^2 < 0$, which is always true for $p > 0$. Intersection points must be real. Discriminant of $(u^2 - 1) x^2 + 4 p u^2 x - u^2 p^2 (u^2 - 1) = 0$. $D = (4 p u^2)^2 - 4 (u^2 - 1) (- u^2 p^2 (u^2 - 1))$ $D = 16 p^2 u^4 + 4 u^2 p^2 (u^2 - 1)^2$ Since $u^2 > 1$, $D > 0$, so real roots exist. One root is positive, one negative. We took the positive one $x_0$. Everything seems consistent.

Final Answer seems to be 100.

One detail: "intersection point of the tangent line to the parabola at P1 and the tangent line to the parabola at P2". Parabola is $y^2 = 4px$. Tangent at $(x_1, y_1)$ is $y y_1 = 2p(x + x_1)$. We found intersection is $(-x_0, 0)$. This is a standard property of parabolas. The intersection of tangents at the endpoints of a focal chord lies on the directrix. Wait, is $P_1 P_2$ a focal chord? The chord passes through $(x_0, y_0)$ and $(x_0, -y_0)$. The line segment is vertical $x = x_0$. Does it pass through the focus $(p, 0)$? Only if $x_0 = p$. But we found $x_0 = 3p$. So it is not a focal chord. The directrix is $x = -p$. The intersection of tangents is at $x = -x_0 = -3p$. So $A$ is at $(-3p, 0)$. This is consistent with our derivation. The property about the directrix applies to focal chords. Since this is not a focal chord, the intersection is not on the directrix. However, the intersection of tangents at points with the same x-coordinate (a chord perpendicular to the axis of symmetry) always lies on the axis of symmetry. Here $P_1$ and $P_2$ share $x_0$. So the chord is vertical. Tangents intersect on the axis of symmetry (x-axis). We found $A = (-x_0, 0)$. This is correct.

Let's re-verify the angle calculation. $A = (-x_0, 0)$, $P_1 = (x_0, y_0)$. Vector $AP_1 = (2x_0, y_0)$. Vector $AP_2 = (2x_0, -y_0)$. Cosine of angle $\theta = 60^\circ$: $\cos 60^\circ = \frac{AP_1 \cdot AP_2}{|AP_1| |AP_2|}$. $AP_1 \cdot AP_2 = 4x_0^2 - y_0^2$. $|AP_1| = \sqrt{4x_0^2 + y_0^2}$. $|AP_2| = \sqrt{4x_0^2 + y_0^2}$. So $\frac{1}{2} = \frac{4x_0^2 - y_0^2}{4x_0^2 + y_0^2}$. $4x_0^2 + y_0^2 = 2(4x_0^2 - y_0^2) = 8x_0^2 - 2y_0^2$. $3y_0^2 = 4x_0^2$. This matches my slope calculation $\tan 30^\circ = y_0 / (2x_0) \implies 1/3 = y_0^2 / (4x_0^2) \implies 4x_0^2 = 3y_0^2$. Correct.

Then $y_0^2 = 4px_0$. $3(4px_0) = 4x_0^2 \implies 3p = x_0$. Correct.

Equation substitution: $(a^2 - p^2) (3p)^2 + 4p a^2 (3p) - a^2 (a^2 - p^2) = 0$. $9p^2 a^2 - 9p^4 + 12p^2 a^2 - a^4 + a^2 p^2 = 0$. $-a^4 + 22 a^2 p^2 - 9p^4 = 0$. $u^4 - 22u^2 + 9 = 0$. Correct.

Solution $u^2 = 11 + 4\sqrt{7}$. $u = \sqrt{7} + 2$. $u^3 = 50 + 19\sqrt{7}$. Floor is 100.

Is there any ambiguity in "one focus of this ellipse coincides with the focus of the parabola"? Parabola focus is $(p, 0)$. Ellipse foci are $(\pm c, 0)$. Could it be $(-c, 0) = (p, 0)$? $-c = p$. Since $c = \sqrt{a^2 - b^2} > 0$ and $p > 0$, this is impossible. So it must be $(c, 0) = (p, 0)$. Correct.

Is there any ambiguity in "intersection point of the tangent line"? No, standard tangents.

Is there any ambiguity in "angle P1 A P2"? No, standard angle.

Is there any ambiguity in "greatest integer"? No, floor function.

The value is uniquely determined.

Let's do a quick check on the value of $19\sqrt{7}$. $19^2 = 361$. $361 \times 7 = 2527$. $\sqrt{2527}$. $50^2 = 2500$. $51^2 = 2601$. So $\sqrt{2527}$ is slightly larger than 50. $50.2...$ So $50 + 50.2... = 100.2...$ Floor is 100.

The calculation is robust.

One final check on the problem statement details. $0 < b < a$. $p > 0$. $\angle P_1 A P_2 = 60^\circ$. Everything used.

Could $u^2 = 11 - 4\sqrt{7}$ be a valid solution? $11 - 4\sqrt{7} \approx 0.42$. $u^2 = a^2/p^2$. $b^2 = a^2 - p^2 = p^2(u^2 - 1)$. If $u^2 < 1$, then $b^2 < 0$, which is impossible for real $b$. So $u^2 = 11 + 4\sqrt{7}$ is the only valid solution.

The result is 100.

One small thing: The problem asks for $\lfloor \frac{a^3}{p^3} \rfloor$. This is $\lfloor u^3 \rfloor$. We calculated $u^3 = 50 + 19\sqrt{7}$. Is it possible that due to floating point precision, it's extremely close to an integer? $19\sqrt{7} = \sqrt{2527}$. Is 2527 a perfect square? $50^2 = 2500$. $51^2 = 2601$. No. Is $50 + \sqrt{2527}$ an integer? No. Is it close to 101? $101 - 50 = 51$. $\sqrt{2527} < 51$. So $50 + \sqrt{2527} < 101$. Is it close to 100? $100 - 50 = 50$. $\sqrt{2527} > 50$. So $50 + \sqrt{2527} > 100$. So it is strictly between 100 and 101. So the floor is exactly 100.

I am confident in the result.

[THOUGHT] The problem asks for $\lfloor a^3/p^3 \rfloor$.

1. Geometric Setup:
   • Ellipse: $x^2/a^2 + y^2/b^2 = 1$. Foci at $(\pm c, 0)$ with $c^2 = a^2 - b^2$.
   • Parabola: $y^2 = 4px$. Focus at $(p, 0)$.
   • Coinciding focus: $c = p \implies a^2 - b^2 = p^2 \implies b^2 = a^2 - p^2$.
2. Intersection Points:
   • Substitute $y^2 = 4px$ into ellipse eq: $(a^2 - p^2)x^2 + 4pa^2x - a^2(a^2 - p^2) = 0$.
   • Let intersection x-coordinate be $x_0$.
3. Tangents Intersection:
   • Tangents to parabola at $(x_0, \pm y_0)$ intersect at $A(-x_0, 0)$.
4. Angle Condition:
   • $\angle P_1 A P_2 = 60^\circ \implies \angle P_1 A X = 30^\circ$.
   • Slope of $AP_1$ is $\tan 30^\circ = 1/\sqrt{3}$.
   • Slope is $y_0 / (2x_0)$.
   • $y_0 / (2x_0) = 1/\sqrt{3} \implies 3y_0^2 = 4x_0^2$.
   • Using $y_0^2 = 4px_0$, we get $12px_0 = 4x_0^2 \implies x_0 = 3p$.
5. Solving for $a/p$:
   • Substitute $x_0 = 3p$ into the quadratic intersection equation.
   • Resulting equation for $u = a/p$: $u^4 - 22u^2 + 9 = 0$.
   • Solutions for $u^2$: $11 \pm 4\sqrt{7}$.
   • Condition $b^2 > 0 \implies u^2 > 1$, so $u^2 = 11 + 4\sqrt{7}$.
6. Calculate Value:
   • $u = \sqrt{11 + 4\sqrt{7}} = 2 + \sqrt{7}$.
   • $u^3 = (2 + \sqrt{7})^3 = 50 + 19\sqrt{7}$.
   • $19\sqrt{7} = \sqrt{2527}$. Since $50^2 = 2500 < 2527 < 2601 = 51^2$, $50 < 19\sqrt{7} < 51$.
   • $100 < 50 + 19\sqrt{7} < 101$.
   • Floor is 100.

[/THOUGHT] \boxed{100}

The value of (\frac{a^3}{p^3}) is (50 + 19\sqrt{7}). Since (19\sqrt{7} = \sqrt{2527}) and (50^2 = 2500 < 2527 < 2601 = 51^2), we have (50 < 19\sqrt{7} < 51). Thus, (100 < 50 + 19\sqrt{7} < 101). The greatest integer not exceeding this value is 100.

\boxed{100}