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This is a Korean CSAT mathematics multiple-choice problem. Solve it and put only the correct choice symbol, one of ①, ②, ③, ④, ⑤, inside \boxed{}.
Problem 26 Letabe a positive real number, and consider the hyperbola x2 a2 − y2 a2 =−1, whose foci areFandF ′. LetPbe the point where the tangent to this hyperbola at (a, √ 2a) meets they-axis. Suppose that P F× P F ′ = 8. Find the value ofa. [3 points] 17 ① √ 3 ② 4 √ 3 3 ③ 5 √ 3 3 ④2 √ 3 ⑤ 7 √ 3 3
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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 26 Letabe a positive real number, and consider the hyperbola x2 a2 − y2 a2 =−1, whose foci areFandF ′. LetPbe the point where the tangent to this hyperbola at (a, √ 2a) meets they-axis. Suppose that P F× P F ′ = 8. Find the value ofa. [3 points] 17 ① √ 3 ② 4 √ 3 3 ③ 5 √ 3 3 ④2 √ 3 ⑤ 7 √ 3 3 더보기
The hyperbola is given by
𝑥 2 𝑎 2 − 𝑦 2 𝑎 2
− 1 , a 2 x 2
− a 2 y 2
=−1,
which is equivalent to
𝑦 2 𝑎 2 − 𝑥 2 𝑎 2
a 2 y 2
− a 2 x 2
=1.
Thus it is a rectangular hyperbola with 𝑏
𝑎 b=a. Its foci are at
𝐹
( 0 , 𝑎 2 ) , 𝐹 ′
( 0 , − 𝑎 2 ) . F=(0,a 2
),F ′ =(0,−a 2
).
The point ( 𝑎 , 2 𝑎 ) (a, 2
a) lies on the hyperbola. The tangent to the hyperbola at ( 𝑥 0 , 𝑦 0 ) (x 0
,y 0
) is
𝑦 𝑦 0 𝑎 2 − 𝑥 𝑥 0 𝑎 2
a 2 yy 0
− a 2 xx 0
=1.
Substituting ( 𝑥 0 , 𝑦 0 )
( 𝑎 , 2 𝑎 ) (x 0
,y 0
)=(a, 2
a) gives
2 𝑎 𝑦 𝑎 2 − 𝑎 𝑥 𝑎 2
1 ⟹ 2 𝑦 − 𝑥
𝑎 . a 2 2
ay
− a 2 ax
=1⟹ 2
y−x=a.
This tangent meets the 𝑦 y-axis ( 𝑥
0 x=0) at
𝑃
( 0 , 𝑎 2 ) . P=(0, 2
a
).
Now compute the distances from 𝑃 P to the foci:
𝑃 𝐹
∣ 𝑎 2 − 𝑎 2 ∣
𝑎 ∣ 1 2 − 2 ∣
𝑎 ⋅ 1 2
𝑎 2 , PF=
2
a
−a 2
=a
2
1
− 2
=a⋅ 2
1
= 2
a
, 𝑃 𝐹 ′
∣ 𝑎 2 − ( − 𝑎 2 ) ∣
𝑎 ( 1 2 + 2 )
𝑎 ⋅ 3 2
3 𝑎 2 . PF ′
2
a
−(−a 2
)
=a( 2
1
2
)=a⋅ 2
3
= 2
3a
.
Their product is
𝑃 𝐹 × 𝑃 𝐹 ′
𝑎 2 ⋅ 3 𝑎 2
3 𝑎 2 2 . PF×PF ′
2
a
⋅ 2
3a
= 2 3a 2
.
We are given 𝑃 𝐹 × 𝑃 𝐹 ′
8 PF×PF ′ =8, so
3 𝑎 2 2
8 ⟹ 3 𝑎 2
16 ⟹ 𝑎 2
16 3 . 2 3a 2
=8⟹3a 2 =16⟹a 2
3 16
.
Since 𝑎
0 a>0,
𝑎
4 3 . a= 3
4
. ② ②
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Comparative Analysis (P41) (Average across runs)
Averages use reported measurements only. Missing usage is not treated as zero.