Problem Analysis #48

The problem asks for $k^2$ where $k$ is the length of segment $HF$. The points are $F(p,0)$, $A(x_A, y_A)$, $H(-p, y_A)$. $A$ is on the parabola $y^2 = 4px$. $H$ is the projection of $A$ on $x=-p$. The ellipse passes through $F, A, H$ with foci on the x-axis.

[PYTHON] import sympy from sympy import symbols, solve, sqrt, Abs

def solve_problem(): p, xA = symbols('p xA', real=True, positive=True)

# yA is positive (1st quadrant)
yA = sqrt(4 * p * xA)

# Area of triangle AHB
# Base AH is horizontal. x-coords are -p and xA. Length = xA - (-p) = xA + p.
# Height is yA.
# Area = 0.5 * (xA + p) * yA
area_eq = 0.5 * (xA + p) * yA - (2 * p + 12)

# Ellipse parameters
# Center xc is midpoint of x-coords of A and H (since they have same y and are on ellipse)
xc = (xA - p) / 2

# F(p, 0) is on the ellipse. Since foci are on x-axis, vertices are on x-axis.
# The y-coord of F is 0, so F is a vertex.
# Semi-major axis a is distance from center to vertex F.
a = Abs(p - xc)

# Equation of ellipse: (x - xc)^2 / a^2 + y^2 / b^2 = 1
# Point A(xA, yA) is on it.
# (xA - xc)^2 / a^2 + yA^2 / b^2 = 1
# Solve for b^2
# term1 = (xA - xc)**2 / a**2
# yA^2 / b^2 = 1 - term1
# b^2 = yA^2 / (1 - term1)

term1 = ((xA - xc)**2) / (a**2)
b2 = yA**2 / (1 - term1)

# Focal distance c_foc
# c_foc^2 = a^2 - b^2
c_foc2 = a**2 - b2
c_foc = sqrt(c_foc2)

# Foci coordinates: (xc - c_foc, 0) and (xc + c_foc, 0)
# B is the focus with positive x-coordinate.
# We need to check which one is positive or if both are, which one is B.
# The problem says "B be the focus ... with positive x-coordinate".
# Let's assume there is a unique such focus or we check conditions later.
# Let's define Bx1 and Bx2.

Bx1 = xc - c_foc
Bx2 = xc + c_foc

# Perimeter equation: AH + AB + HB = p + 27
# AH = xA + p
# We need to solve the system for p and xA.
# Since symbolic solution might be hard with Abs and sqrt, let's try to simplify or use numerical solver if needed.
# But first, let's try to express everything in terms of p and xA and see if we can simplify.

# Let's print the expressions to see complexity
print(f"Area Eq: {area_eq}")
print(f"xc: {xc}")
print(f"a: {a}")
print(f"b2: {b2}")

solve_problem() [/PYTHON]

The problem asks for $k^2$, where $k$ is the length of the segment $HF$. The parabola is defined by $y^2 = 4px$ with focus $F(p, 0)$ and directrix $x = -p$. Point $A(x_A, y_A)$ is on the parabola in the first quadrant, so $y_A = \sqrt{4px_A}$. Point $H$ is the projection of $A$ onto the directrix, so $H(-p, y_A)$. The ellipse passes through $F, A, H$ with foci on the $x$-axis. Let the center of the ellipse be $(x_c, 0)$. Since $A$ and $H$ are on the ellipse and share the same $y$-coordinate, and the ellipse is symmetric about its major axis (the $x$-axis) and its minor axis (vertical line $x=x_c$), the center's $x$-coordinate must be the midpoint of the $x$-coordinates of $A$ and $H$. $x_c = \frac{x_A + (-p)}{2} = \frac{x_A - p}{2}$. Since $F(p, 0)$ is on the ellipse and on the major axis, $F$ is a vertex. The semi-major axis $a$ is the distance from the center to the vertex: $a = |p - x_c| = |p - \frac{x_A - p}{2}| = |\frac{2p - x_A + p}{2}| = \frac{3p - x_A}{2}$ (assuming $3p > x_A$, which turns out to be true).

Let $B$ be a focus of the ellipse. The sum of distances from any point on the ellipse to the two foci is $2a$. Due to the symmetry of $A$ and $H$ with respect to the vertical line $x=x_c$, and the symmetry of the foci with respect to the same line, the distance from $H$ to focus $B$ is equal to the distance from $A$ to the other focus $B'$. Thus, $AB + HB = AB + AB' = 2a$. The perimeter of triangle $AHB$ is $AH + AB + HB = AH + 2a$. $AH$ is the horizontal distance between $H(-p, y_A)$ and $A(x_A, y_A)$, so $AH = x_A - (-p) = x_A + p$. Perimeter $= (x_A + p) + (3p - x_A) = 4p$. Given perimeter is $p + 27$, we have: $4p = p + 27 \implies 3p = 27 \implies p = 9$.

The area of triangle $AHB$ is $\frac{1}{2} \times \text{base} \times \text{height}$. Base $AH = x_A + p$. Height $= y_A = \sqrt{4px_A} = 2\sqrt{px_A}$. Area $= \frac{1}{2}(x_A + p)(2\sqrt{px_A}) = (x_A + p)\sqrt{px_A}$. Given area is $2p + 12$. Substituting $p=9$: $(x_A + 9)\sqrt{9x_A} = 2(9) + 12 = 30$. $3\sqrt{x_A}(x_A + 9) = 30$. $\sqrt{x_A}(x_A + 9) = 10$. Let $u = \sqrt{x_A}$. $u(u^2 + 9) = 10 \implies u^3 + 9u - 10 = 0$. $(u-1)(u^2 + u + 10) = 0$. The only real positive solution is $u=1$, so $x_A = 1$.

We need to compute $k^2 = HF^2$. $H = (-p, y_A) = (-9, 2\sqrt{9 \cdot 1}) = (-9, 6)$. $F = (p, 0) = (9, 0)$. $k^2 = (9 - (-9))^2 + (0 - 6)^2 = 18^2 + (-6)^2 = 324 + 36 = 360$.

[THOUGHT] The calculation is verified. p = 9 xA = 1 H = (-9, 6) F = (9, 0) k^2 = 18^2 + 6^2 = 324 + 36 = 360. [/THOUGHT]

\boxed{360}