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] Derive intersection coordinates using the 60° condition, then use the ellipse/parabola intersection and shared-focus condition (p^2=a^2-b^2) to solve for the ratio a/p, and finally compute floor((a/p)^3) exactly with SymPy. [/THOUGHT] [PYTHON] import sympy as sp
sqrt7 = sp.sqrt(7) r = (1 + 2*sqrt7)/6 # r = a^2/b^2 s = sp.simplify(r/(r-1)) # (a/p)^2 expr = sp.simplify(s**sp.Rational(3,2)) # (a/p)^3 s_simplified = sp.nsimplify(s) expr_n = sp.N(expr, 50) print("r =", r) print("s = (a/p)^2 =", sp.simplify(s)) print("numeric s =", sp.N(s, 50)) print("(a/p)^3 =", expr) print("numeric (a/p)^3 =", expr_n) print("floor =", int(sp.floor(expr)))
also show exact rationalized form
print("s rationalized =", sp.simplify(sp.together(s))) print("s rationalized2 =", sp.simplify((1+2sqrt7)/(2sqrt7-5))) print("expr simplified radical =", sp.simplify(expr.rewrite(sp.sqrt))) print("expr minimal polynomial? degree:", sp.minpoly(expr).degree()) print("minpoly:", sp.minpoly(expr)) print("expr as root:", sp.nsimplify(expr, [sp.sqrt(7)]))
confirm inequality around integer boundary
val = sp.N(expr, 80) print("val-101 =", val-101) print("val-102 =", val-102) [/PYTHON][THOUGHT] Derive intersection coordinates using the 60° condition, then use the ellipse/parabola intersection and shared-focus condition (p^2=a^2-b^2) to solve for the ratio a/p, and finally compute floor((a/p)^3) exactly with SymPy. [/THOUGHT] [PYTHON] import sympy as sp
sqrt7 = sp.sqrt(7) r = (1 + 2*sqrt7)/6 # r = a^2/b^2 s = sp.simplify(r/(r-1)) # (a/p)^2 expr = sp.simplify(s**sp.Rational(3,2)) # (a/p)^3
print("r =", r) print("s = (a/p)^2 =", s) print("numeric s =", sp.N(s, 50)) print("(a/p)^3 =", expr) print("numeric (a/p)^3 =", sp.N(expr, 80)) print("floor =", sp.floor(expr)) print("val-101 =", sp.N(expr-101, 80)) print("val-102 =", sp.N(expr-102, 80)) [/PYTHON]
[THOUGHT] From the verified computation, ((a/p)^3=(4\sqrt7+11)^{3/2}\approx 100.26927), so (\left\lfloor a^3/p^3\right\rfloor=100). \boxed{100}