Consider the ellipseand suppose that one focus of this ellipse coincides with the focus of the parabolaLet the two intersection points of the parabola and the ellipse be denoted by and .Let be the intersection point of the tangent line to the parabola at and the tangent line to the parabola at .Given that the angle , and that for any real number , denotes the greatest integer not exceeding , determine the value ofoindent
[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