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로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.

For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.

Problem 29 Consider the parabola with focusF(p,0) (p >0) and directrixx=−p, as shown in the figure. Among the points on this parabola, letAbe a point in the first quadrant, and let Hbe the foot of the perpendicular fromAto the directrix. Consider the ellipse whose foci lie on thex-axis and which passes through the three pointsF,A, andH. LetBbe the focus of this ellipse with positivex-coordinate. Suppose that triangleAHBhas perimeter p+ 27 and area 2p+ 12. Let the length of segment HFbek. Computek 2. [4 points] 더보기

Let’s reason carefully through the problem.

We have a parabola with focus 𝐹 ( 𝑝 , 0 ) F(p,0) ( 𝑝

0 p>0) and directrix 𝑥

− 𝑝 x=−p. Its equation is 𝑦 2

4 𝑝 𝑥 y 2 =4px. Point 𝐴 A is in the first quadrant on the parabola, so 𝐴

( 𝑥 , 𝑦 ) A=(x,y) with 𝑦

0 y>0, 𝑥

0 x>0, and 𝑦 2

4 𝑝 𝑥 y 2 =4px.

𝐻 H is the foot of the perpendicular from 𝐴 A to the directrix, so 𝐻

( − 𝑝 , 𝑦 ) H=(−p,y).

An ellipse has its foci on the 𝑥 x-axis and passes through 𝐹 F, 𝐴 A, and 𝐻 H. Let its foci be ( 𝑐 , 0 ) (c,0) and ( − 𝑐 , 0 ) (−c,0) with 𝑐

0 c>0. 𝐵 B is the focus with positive 𝑥 x-coordinate, so 𝐵

( 𝑐 , 0 ) B=(c,0).

For any point 𝑃 P on the ellipse, the sum of distances to the foci is constant, say 2 𝑎 2a. For 𝐹 ( 𝑝 , 0 ) F(p,0): ∣ 𝑝 − 𝑐 ∣ + ( 𝑝 + 𝑐 )

2 𝑎 ∣p−c∣+(p+c)=2a. If 𝑝 ≥ 𝑐 p≥c, then 2 𝑎

2 𝑝 2a=2p so 𝑎

𝑝 a=p. If 𝑝 < 𝑐 p<c, then 2 𝑎

2 𝑐 2a=2c so 𝑎

𝑐 a=c.

For 𝐻 ( − 𝑝 , 𝑦 ) H(−p,y): ( 𝑐 + 𝑝 ) 2 + 𝑦 2 + ( 𝑐 − 𝑝 ) 2 + 𝑦 2

2 𝑎 (c+p) 2 +y 2 ​

(c−p) 2 +y 2 ​

=2a. If 𝑝 ≥ 𝑐 p≥c, the left side is ≥ ( 𝑐 + 𝑝 ) + ( 𝑝 − 𝑐 )

2 𝑝 ≥(c+p)+(p−c)=2p with equality only if 𝑦

0 y=0. Since 𝑦

0 y>0, the sum is

2 𝑝

2p, contradicting 2 𝑎 = 2 𝑝 2a=2p. If 𝑝 < 𝑐 p<c, then 2 𝑎 = 2 𝑐 2a=2c, and the left side is ≥ ( 𝑐

𝑝 ) + ( 𝑐 − 𝑝 )

2 𝑐 ≥(c+p)+(c−p)=2c with equality only if 𝑦

0 y=0. Again 𝑦

0 y>0 gives a contradiction.

Thus no non‑degenerate ellipse exists under the given conditions. The problem statement is inconsistent.

Since the problem asks to compute 𝑘 2 k 2 where 𝑘

𝐻 𝐹 k=HF, and no valid configuration exists, 𝑘 2 k 2 cannot be determined.

\boxed{\text{No solution exists under the given conditions.}}

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