[THOUGHT] We need to parse the geometry carefully.

Given:

• Parabola: focus F(p,0), directrix x = -p. Standard parabola: set of points equidistant from focus and directrix. For any point A(x_A, y_A) on parabola: distance to F = distance to directrix => sqrt((x_A - p)^2 + y_A^2) = x_A + p (since distance to line x = -p is x_A - (-p) = x_A + p). Squaring: (x_A - p)^2 + y_A^2 = (x_A + p)^2 => x_A^2 - 2p x_A + p^2 + y_A^2 = x_A^2 + 2p x_A + p^2 => y_A^2 = 4p x_A. So parabola equation: y^2 = 4p x, x ≥ 0.

• A is in first quadrant, so x_A > 0, y_A > 0.

• H is foot of perpendicular from A to directrix x = -p. So H = (-p, y_A).

• Ellipse with foci on x-axis, passing through F, A, H. Let the foci be (c1, 0) and (c2, 0). Since ellipse passes through F(p,0) and H(-p, y_A) and A(x_A, y_A). The foci are on x-axis, so ellipse is horizontal major axis.

• Let B be the focus with positive x-coordinate. So B = (c, 0) with c > 0. The other focus is at (-c, 0) perhaps? Wait: "foci lie on the x-axis" but not necessarily symmetric about origin. However, ellipse passes through F(p,0) and H(-p, y_A). For an ellipse with foci (f1,0) and (f2,0), the sum of distances from any point on ellipse to the foci is constant = 2a (major axis length). So we have:

For point F(p,0): |p - f1| + |p - f2| = 2a. For point H(-p, y_A): sqrt((-p - f1)^2 + y_A^2) + sqrt((-p - f2)^2 + y_A^2) = 2a. For point A(x_A, y_A): sqrt((x_A - f1)^2 + y_A^2) + sqrt((x_A - f2)^2 + y_A^2) = 2a.

We also know A is related to p: y_A^2 = 4p x_A.

We are told B is the focus with positive x-coordinate. So B = (c, 0) with c > 0. The other focus is somewhere else (maybe negative). Let's denote the two foci as (f, 0) and (g, 0). Without loss, let f = c (positive) and g = d (could be negative). Then B = (c, 0).

Now triangle AHB: vertices A(x_A, y_A), H(-p, y_A), B(c, 0). Perimeter = p + 27, area = 2p + 12.

We need length HF = distance between H(-p, y_A) and F(p,0). That's HF = sqrt((2p)^2 + y_A^2) = sqrt(4p^2 + y_A^2). Let k = HF, compute k^2 = 4p^2 + y_A^2.

We need to find p and y_A (and maybe x_A, c) from conditions.

Let's set up equations.

Given parabola: y_A^2 = 4p x_A.

Coordinates: A = (x_A, y_A) H = (-p, y_A) F = (p, 0) B = (c, 0) with c > 0.

Triangle AHB: Sides: AH = distance between A and H = sqrt((x_A + p)^2 + (y_A - y_A)^2) = x_A + p. HB = distance between H and B = sqrt((c + p)^2 + y_A^2). AB = distance between A and B = sqrt((x_A - c)^2 + y_A^2).

Perimeter: AH + HB + AB = (x_A + p) + sqrt((c + p)^2 + y_A^2) + sqrt((x_A - c)^2 + y_A^2) = p + 27.

Area: triangle vertices A(x_A, y_A), H(-p, y_A), B(c, 0). Base AH is horizontal length x_A + p. Height from B to line AH: line AH is horizontal at y = y_A, so vertical distance from B to line AH is |y_A - 0| = y_A. So area = 1/2 * base * height = 1/2 * (x_A + p) * y_A = 2p + 12.

Thus we have two equations:

(1) (x_A + p) * y_A = 2*(2p + 12) = 4p + 24.

(2) (x_A + p) + sqrt((c + p)^2 + y_A^2) + sqrt((x_A - c)^2 + y_A^2) = p + 27.

Also, ellipse condition: sum of distances from any point on ellipse to foci (c,0) and (d,0) is constant = 2a. But we don't know d. However, ellipse passes through F(p,0) and H(-p, y_A). So for F: |p - c| + |p - d| = 2a. For H: sqrt((-p - c)^2 + y_A^2) + sqrt((-p - d)^2 + y_A^2) = 2a. For A: sqrt((x_A - c)^2 + y_A^2) + sqrt((x_A - d)^2 + y_A^2) = 2a.

We have three unknowns: x_A, y_A, c, d, p. But p is also unknown. Actually p is given as a parameter in the parabola definition, but we need to find p from the conditions. So unknowns: p, x_A, y_A, c, d. Equations:

(0) y_A^2 = 4p x_A. (1) (x_A + p) * y_A = 4p + 24. (2) (x_A + p) + sqrt((c + p)^2 + y_A^2) + sqrt((x_A - c)^2 + y_A^2) = p + 27. (3) ellipse condition: sum of distances from F to foci = sum of distances from H to foci = sum of distances from A to foci.

Let's denote S = 2a = constant sum.

For F: |p - c| + |p - d| = S. For H: sqrt((-p - c)^2 + y_A^2) + sqrt((-p - d)^2 + y_A^2) = S. For A: sqrt((x_A - c)^2 + y_A^2) + sqrt((x_A - d)^2 + y_A^2) = S.

We need to find d as well.

But maybe we can deduce something: Since ellipse passes through F and H, and foci are on x-axis, maybe the other focus is symmetric? Not necessarily.

However, note that H and F are symmetric about the y-axis? F is (p,0), H is (-p, y_A). Not symmetric unless y_A = 0. So not.

We might try to eliminate d. Let's denote f1 = c, f2 = d.

From F: |p - c| + |p - d| = S. Since c > 0, d could be negative or positive.

We don't know signs. Let's assume d < 0 perhaps? Not sure.

But maybe we can use the fact that ellipse passes through A and H, and A is on parabola y_A^2 = 4p x_A.

Alternatively, maybe we can find c from triangle perimeter equation and area equation, then find p.

Let's try to solve for p, x_A, y_A from (0) and (1). Then we can find c from (2) maybe.

But we also have ellipse condition linking c and d. However, we might not need d if we can find c from triangle and ellipse condition separately? Wait, we need ellipse to pass through F, A, H. That gives three equations for S, c, d. But we also have triangle condition involving c. So we have more equations than unknowns, so p may be determined uniquely.

Let's attempt to solve numerically using Python.

We have unknowns: p, x_A, y_A, c, d.

Equations: eq1: y_A**2 - 4px_A = 0. eq2: (x_A + p) * y_A - (4*p + 24) = 0. eq3: (x_A + p) + sqrt((c + p)2 + y_A2) + sqrt((x_A - c)2 + y_A2) - (p + 27) = 0. eq4: (|p - c| + |p - d|) - (sqrt((-p - c)2 + y_A2) + sqrt((-p - d)2 + y_A2)) = 0. eq5: (sqrt((x_A - c)2 + y_A2) + sqrt((x_A - d)2 + y_A2)) - (sqrt((-p - c)2 + y_A2) + sqrt((-p - d)2 + y_A2)) = 0.

But eq4 and eq5 are essentially saying sums are equal. Actually we need all three sums equal to S. So we can set:

S1 = |p - c| + |p - d| S2 = sqrt((-p - c)2 + y_A2) + sqrt((-p - d)2 + y_A2) S3 = sqrt((x_A - c)2 + y_A2) + sqrt((x_A - d)2 + y_A2)

We require S1 = S2 = S3.

So we have two equations: S1 - S2 = 0, S2 - S3 = 0.

But absolute values depend on signs of p-c, p-d. We need to consider cases.

Given c > 0. p > 0. So p - c could be positive or negative depending on c. p - d sign unknown.

Let's try to guess signs: Since ellipse passes through F(p,0) and H(-p, y_A), maybe foci are symmetric about origin? Not necessarily. But maybe the ellipse is centered at origin? If foci are symmetric about origin, then d = -c. Then B = (c,0) and other focus (-c,0). Let's test that hypothesis.

Assume d = -c. Then foci are (c,0) and (-c,0). Then S = |p - c| + |p + c| = (p - c if c < p else c - p) + (p + c) = ? Let's compute: If c < p, then |p - c| = p - c, |p + c| = p + c, sum = 2p. If c > p, then |p - c| = c - p, |p + c| = p + c, sum = 2c. So S = 2 max(p, c). But also S must equal sum for H: sqrt((-p - c)^2 + y_A^2) + sqrt((-p + c)^2 + y_A^2) = sqrt((p + c)^2 + y_A^2) + sqrt((p - c)^2 + y_A^2). And for A: sqrt((x_A - c)^2 + y_A^2) + sqrt((x_A + c)^2 + y_A^2).

We can test if this works.

But we don't know if d = -c. Let's not assume.

Alternatively, maybe we can eliminate d by subtracting equations. Let's define functions.

Better to use numerical solving with sympy or scipy.

We'll try to solve the system assuming c > 0, d real.

We'll need to handle absolute values. Let's consider case: c < p and d < p? Not sure.

Given p > 0, c > 0. Could be c < p or c > p. Similarly d could be negative.

Given ellipse passes through H(-p, y_A) which is left of origin, maybe the other focus is also left of origin? Possibly d < -p? Not sure.

Let's attempt to solve numerically using fsolve from scipy.optimize.

We'll treat variables: p, x_A, y_A, c, d.

But note: y_A^2 = 4p x_A, so we can eliminate x_A: x_A = y_A^2/(4p). Then eq2 becomes (y_A^2/(4p) + p) * y_A = 4p + 24.

Simplify: y_A * (y_A^2/(4p) + p) = 4p + 24 => (y_A^3)/(4p) + p y_A = 4p + 24.

Multiply by 4p: y_A^3 + 4p^2 y_A = 16p^2 + 96p.

Thus eq2': y_A^3 + 4p^2 y_A - 16p^2 - 96p = 0.

Now eq3: (x_A + p) + sqrt((c + p)^2 + y_A^2) + sqrt((x_A - c)^2 + y_A^2) = p + 27.

Substitute x_A = y_A^2/(4p). Then x_A + p = y_A^2/(4p) + p.

So eq3: y_A^2/(4p) + p + sqrt((c + p)^2 + y_A^2) + sqrt((y_A^2/(4p) - c)^2 + y_A^2) = p + 27.

Cancel p on both sides? Actually left has +p, right has p+27, so subtract p both sides: y_A^2/(4p) + sqrt((c + p)^2 + y_A^2) + sqrt((y_A^2/(4p) - c)^2 + y_A^2) = 27.

Thus eq3': y_A^2/(4p) + sqrt((c + p)^2 + y_A^2) + sqrt((y_A^2/(4p) - c)^2 + y_A^2) - 27 = 0.

Now eq4: S1 = |p - c| + |p - d|. S2 = sqrt((-p - c)^2 + y_A^2) + sqrt((-p - d)^2 + y_A^2) = sqrt((p + c)^2 + y_A^2) + sqrt((p + d)^2 + y_A^2). S3 = sqrt((x_A - c)^2 + y_A^2) + sqrt((x_A - d)^2 + y_A^2).

We require S1 = S2 = S3.

Thus two equations: S1 - S2 = 0, S2 - S3 = 0.

We need to handle absolute values in S1. Let's consider possible cases for p - c and p - d.

Case 1: c < p, d < p. Then |p - c| = p - c, |p - d| = p - d, so S1 = 2p - (c + d).

Case 2: c < p, d > p. Then |p - c| = p - c, |p - d| = d - p, S1 = d - c.

Case 3: c > p, d < p. Then S1 = c - d.

Case 4: c > p, d > p. Then S1 = (c - p) + (d - p) = c + d - 2p.

We don't know.

Given ellipse passes through H(-p, y_A) which is left of origin, maybe one focus is left, one right. Possibly d < -p? Not sure.

We can try to solve numerically without specifying case, by using absolute values in Python's abs function.

Let's write code to solve the system of equations using sympy or scipy.

We'll use sympy's nsolve.

We have variables: p, y_A, c, d. (x_A eliminated).

Equations: eq2': y_A3 + 4*p2y_A - 16p2 - 96*p = 0. eq3': y_A2/(4p) + sqrt((c + p)2 + y_A2) + sqrt((y_A**2/(4p) - c)2 + y_A2) - 27 = 0. eq4': abs(p - c) + abs(p - d) - (sqrt((p + c)2 + y_A2) + sqrt((p + d)2 + y_A2)) = 0. eq5': (sqrt((p + c)2 + y_A2) + sqrt((p + d)2 + y_A2)) - (sqrt((y_A2/(4*p) - c)2 + y_A2) + sqrt((y_A2/(4*p) - d)2 + y_A2)) = 0.

We need initial guesses. Since A is in first quadrant, y_A > 0. p > 0. c > 0. d maybe negative.

Let's try to guess p ~ 10? Let's see if we can get reasonable numbers.

We'll attempt to solve using nsolve.

But note: eq4 and eq5 involve square roots and absolute values, may cause issues. We can try to square or use sympy's piecewise.

Alternatively, we can assume specific sign cases. Let's assume c < p and d < p? Then S1 = 2p - (c + d). Let's try that.

Case: c < p, d < p. Then eq4 becomes (2p - c - d) - (sqrt((p + c)^2 + y_A^2) + sqrt((p + d)^2 + y_A^2)) = 0.

eq5 unchanged.

We can attempt solve.

Let's code.

We'll import sympy and use nsolve.

We need to provide initial guesses. Let's guess p=10, y_A=20, c=5, d=5? But d must be < p, maybe negative? Actually if d < p, could be negative. Let's guess d = -5.

Let's try.

We'll write code. </think>

[PYTHON] import sympy as sp

Define variables

p, yA, c, d = sp.symbols('p yA c d', positive=True) # p>0, yA>0, c>0, but d may be negative; we can treat d as real

We'll not enforce positive for d.

Equation eq2'

eq2 = yA3 + 4*p2yA - 16p**2 - 96*p

Equation eq3'

xA = yA**2/(4*p) eq3 = xA + sp.sqrt((c + p)2 + yA2) + sp.sqrt((xA - c)2 + yA2) - 27

Equation eq4: case c < p and d < p (so absolute values as p-c and p-d)

But we'll use absolute values directly with abs

eq4 = abs(p - c) + abs(p - d) - (sp.sqrt((p + c)2 + yA2) + sp.sqrt((p + d)2 + yA2))

Equation eq5

eq5 = (sp.sqrt((p + c)2 + yA2) + sp.sqrt((p + d)2 + yA2)) - (sp.sqrt((xA - c)2 + yA2) + sp.sqrt((xA - d)2 + yA2))

Initial guesses: p=10, yA=20, c=5, d=-5

initial_guess = [10, 20, 5, -5]

Use nsolve

try: sol = sp.nsolve([eq2, eq3, eq4, eq5], [p, yA, c, d], initial_guess, tol=1e-14, maxsteps=100) print("Solution found:", sol) except Exception as e: print("Error:", e) # Try different initial guesses # Let's try p=15, yA=30, c=10, d=-10 initial_guess2 = [15, 30, 10, -10] try: sol2 = sp.nsolve([eq2, eq3, eq4, eq5], [p, yA, c, d], initial_guess2, tol=1e-14, maxsteps=100) print("Solution found:", sol2) except Exception as e2: print("Error with second guess:", e2) [/PYTHON]

[THOUGHT] From the Python output, we have p ≈ 9.0000000000000000000000065347, yA ≈ 6.0000000000000000000000040241, c ≈ -15.25833, d ≈ 7.25833.

We need k = HF. Recall H is foot from A to directrix x = -p. So A = (xA, yA), H = (-p, yA). F = (p, 0). So HF = distance between (-p, yA) and (p, 0):

HF^2 = (2p)^2 + yA^2 = 4p^2 + yA^2.

From p ≈ 9, yA ≈ 6:

4p^2 + yA^2 = 4*(81) + 36 = 324 + 36 = 360.

But let's compute exactly using the high-precision numbers from the solution.

We can compute 4*p^2 + yA^2 using the given values.

From the Python output: p ≈ 9.0000000000000000000000065347, yA ≈ 6.0000000000000000000000040241.

We'll compute 4*p^2 + yA^2 numerically to see if it's exactly 360 or something else.

Let's do a quick Python check in my head: p^2 ≈ 81.000..., 4p^2 ≈ 324.000..., yA^2 ≈ 36.000..., sum ≈ 360.000... So k^2 = 360.

But we should verify if there's any slight deviation due to floating error. The given numbers are very close to integers 9 and 6, so k^2 should be exactly 360.

Thus final answer: \boxed{360}. </think>

\boxed{360}