Problem Analysis #23

[THOUGHT] To solve this problem, we need to find the relationship between the parameters (a), (b), and (p) given the geometric conditions.

First, we identify the foci of the ellipse. For the ellipse (\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1) with (a > b), the foci are located at ((\pm c, 0)) where (c = \sqrt{a^2 - b^2}).

The focus of the parabola (y^2 = 4px) is at ((p, 0)). According to the problem, one focus of the ellipse coincides with the focus of the parabola. Therefore, (c = p).

Thus, we have: [ p = \sqrt{a^2 - b^2} ] Squaring both sides gives: [ p^2 = a^2 - b^2 ] [ b^2 = a^2 - p^2 ]

Next, we consider the intersection points (\mathrm{P}_1) and (\mathrm{P}_2) of the parabola and the ellipse. Substitute (y^2 = 4px) into the ellipse equation: [ \frac{x^2}{a^2} + \frac{4px}{b^2} = 1 ] Using (b^2 = a^2 - p^2), we substitute: [ \frac{x^2}{a^2} + \frac{4px}{a^2 - p^2} = 1 ] Multiply through by (a^2(a^2 - p^2)) to clear the denominators: [ x^2(a^2 - p^2) + 4pxa^2 = a^2(a^2 - p^2) ] [ x^2a^2 - x^2p^2 + 4pxa^2 = a^4 - a^2p^2 ] [ x^2a^2 + 4pxa^2 - x^2p^2 = a^4 - a^2p^2 ] [ x^2(a^2 - p^2) + 4pxa^2 = a^4 - a^2p^2 ] [ x^2(a^2 - p^2) + 4pxa^2 = a^2(a^2 - p^2) ] Factor out (a^2 - p^2): [ (x^2 + 4px) = a^2 ] This is a quadratic equation in (x): [ x^2 + 4px - a^2 = 0 ] Using the quadratic formula (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}) with (a = 1), (b = 4p), and (c = -a^2), we get: [ x = \frac{-4p \pm \sqrt{(4p)^2 + 4a^2}}{2} ] [ x = \frac{-4p \pm \sqrt{16p^2 + 4a^2}}{2} ] [ x = \frac{-4p \pm 2\sqrt{4p^2 + a^2}}{2} ] [ x = -2p \pm \sqrt{4p^2 + a^2} ]

Thus, the (x)-coordinates of (\mathrm{P}_1) and (\mathrm{P}_2) are: [ x_1 = -2p + \sqrt{4p^2 + a^2} ] [ x_2 = -2p - \sqrt{4p^2 + a^2} ]

The corresponding (y)-coordinates are: [ y_1 = \pm \sqrt{4p(-2p + \sqrt{4p^2 + a^2})} = \pm 2\sqrt{p\sqrt{4p^2 + a^2} - 2p^2} ] [ y_2 = \pm \sqrt{4p(-2p - \sqrt{4p^2 + a^2})} = \pm 2\sqrt{-p\sqrt{4p^2 + a^2} - 2p^2} ]

Since (y_2) is imaginary, we only consider (y_1). Thus, the intersection points are: [ \mathrm{P}_1 = \left(-2p + \sqrt{4p^2 + a^2}, 2\sqrt{p\sqrt{4p^2 + a^2} - 2p^2}\right) ] [ \mathrm{P}_2 = \left(-2p + \sqrt{4p^2 + a^2}, -2\sqrt{p\sqrt{4p^2 + a^2} - 2p^2}\right) ]

The tangent lines to the parabola at (\mathrm{P}_1) and (\mathrm{P}_2) are: [ yy_1 = 2p(x + x_1) ] [ yy_2 = 2p(x + x_2) ]

Substituting (y_1) and (y_2) and simplifying, we find that the tangent lines intersect at the point (A = (0, 0)).

Given (\angle \mathrm{P}_1 A \mathrm{P}_2 = 60^\circ), we use the fact that the angle between the tangents to a parabola at its points of intersection with another conic section is related to the eccentricity of the conic section. For an ellipse, the angle (\theta) between the tangents at the points of intersection with a parabola is given by: [ \cos(\theta) = \frac{b^2}{a^2} ]

Given (\theta = 60^\circ), we have: [ \cos(60^\circ) = \frac{1}{2} ] [ \frac{b^2}{a^2} = \frac{1}{2} ] [ b^2 = \frac{a^2}{2} ]

Substitute (b^2 = a^2 - p^2): [ \frac{a^2}{2} = a^2 - p^2 ] [ p^2 = a^2 - \frac{a^2}{2} ] [ p^2 = \frac{a^2}{2} ] [ p = \frac{a}{\sqrt{2}} ]

Thus, we need to find: [ \left\lfloor \frac{a^3}{p^3} \right\rfloor ] [ p = \frac{a}{\sqrt{2}} \Rightarrow p^3 = \left(\frac{a}{\sqrt{2}}\right)^3 = \frac{a^3}{2\sqrt{2}} ] [ \frac{a^3}{p^3} = \frac{a^3}{\frac{a^3}{2\sqrt{2}}} = 2\sqrt{2} ] [ 2\sqrt{2} \approx 2 \times 1.414 = 2.828 ]

Finally, we take the floor value: [ \left\lfloor 2.828 \right\rfloor = 2 ]

[/THOUGHT] \boxed{2}