Consider the parabola with focus and directrix , as shown in the figure. Among the points on this parabola, let be a point in the first quadrant, and let be the foot of the perpendicular from to the directrix. Consider the ellipse whose foci lie on the -axis and which passes through the three points , , and . Let be the focus of this ellipse with positive -coordinate. Suppose that triangle has perimeter and area . Let the length of segment be . Compute . [4 points]
[THOUGHT] We need to solve for the parameters of the configuration.
Let the parabola be (y^2 = 4px).
For a point (A(x_A,y_A)) on it we have (y_A^2 = 4p x_A).
The foot of the perpendicular from (A) to the directrix (x=-p) is (H(-p,y_A)).
Let the ellipse have foci (B(d,0)) (with (d>0)) and (D(-e,0)) (with (e>0)).
Since the ellipse passes through (F(p,0),A,H), the sum of distances to the foci is constant:
[
\begin{aligned}
S &= |p-d|+p+e = 2p+e-d,\
S &= \sqrt{(d+p)^2+y_A^2}+\sqrt{(e-p)^2+y_A^2},\
S &= \sqrt{(x_A-d)^2+y_A^2}+\sqrt{(x_A+e)^2+y_A^2}.
\end{aligned}
]
The triangle (AHB) has vertices (A(x_A,y_A), H(-p,y_A), B(d,0)).
Its side lengths are:
[
HA = x_A+p,\qquad BH = \sqrt{(d+p)^2+y_A^2},\qquad AB = \sqrt{(x_A-d)^2+y_A^2}.
]
Given: [ HA+AB+BH = p+27\quad\Longrightarrow\quad AB+BH = 27-x_A, ] [ \text{Area} = \frac12,(HA),y_A = 2p+12\quad\Longrightarrow\quad (p+x_A)y_A = 4p+24. ]
Thus we have the system: [ \begin{cases} y_A^2 = 4p x_A,\ (p+x_A)y_A = 4p+24,\ \sqrt{(x_A-d)^2+y_A^2}+\sqrt{(d+p)^2+y_A^2}=27-x_A,\ \sqrt{(d+p)^2+y_A^2}+\sqrt{(e-p)^2+y_A^2}=2p+e-d,\ \sqrt{(x_A-d)^2+y_A^2}+\sqrt{(x_A+e)^2+y_A^2}=2p+e-d. \end{cases} ]
We solve this system for positive real numbers (p,x_A,y_A,d,e).
Then (k = HF = \sqrt{4p^2+y_A^2}) and we need (k^2 = 4p^2+y_A^2).
We will use SymPy to find a solution. [/THOUGHT]
[PYTHON] import sympy as sp
Define symbols
p, x, y, d, e = sp.symbols('p x y d e', positive=True, real=True)
Equations
eq1 = sp.Eq(y**2, 4px) # parabola eq2 = sp.Eq((p + x)y, 4p + 24) # area condition eq3 = sp.Eq(sp.sqrt((x - d)2 + y2) + sp.sqrt((d + p)2 + y2), 27 - x) # perimeter eq4 = sp.Eq(sp.sqrt((d + p)2 + y2) + sp.sqrt((e - p)2 + y2), 2p + e - d) # ellipse F=H eq5 = sp.Eq(sp.sqrt((x - d)2 + y2) + sp.sqrt((x + e)2 + y2), 2p + e - d) # ellipse F=A
Solve numerically
sol = sp.nsolve([eq1, eq2, eq3, eq4, eq5], (p, x, y, d, e), (12, 0.7, 5.7, 2, 15)) sol [/PYTHON]
[THOUGHT] The solution from the Python code gives p=9, x_A=1, y_A=6. The length HF is the distance between H(-p, y_A) and F(p,0), so HF^2 = (2p)^2 + y_A^2 = 4*81 + 36 = 324+36=360. [/THOUGHT]
\boxed{360}