درس الفلكة
اﻟﻔـﻠﻜـــــﺔ
-I
La sphère
اﻟﻔﻠﻜﺔ اﻟﻤﻌﺮﻓﺔ ﺑﻤﺮآﺰهﺎ وﺷﻌﺎﻋﻬﺎ.
ﺗﻌﺮﻳﻒ :
ﻟﺘﻜﻦ Aﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻀﺎء ξو Rﻋﺪد ﺣﻘﻴﻘﻲ.
اﻟﻔﻠﻜﺔ اﻟﺘﻲ ﻣﺮآﺰهﺎ Aوﺷﻌﺎﻋﻬﺎ Rهﻲ ﻣﺠﻤﻮﻋﺔ اﻟﻨﻘﻂ Mﺣﻴﺚ :
) S ( A, R
وﻧﺮﻣﺰ ﻟﻬﺎ ﺑــ :
. AM = R
}S ( A, R ) = {M ∈ ξ / AM = R
ﻣﻌﺎدﻟﺔ ﻓﻠﻜﺔ :
(1ﻣﻌﺎدﻟﺔ ﻓﻠﻜﺔ ﻣﻌﺮﻓﺔ ﺑﻤﺮآﺰهﺎ وﺷﻌﺎﻋﻬﺎ :
اﻟﻔﻀﺎء ξﻣﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ . O, i , j , k
(
)
ﻟﺘﻜﻦ S ( Ω, R ) :ﻓﻠﻜﺔ ﻣﺮآﺰهﺎ ) Ω ( x0 , y0 , z0وﺷﻌﺎﻋﻬﺎ Rﺣﻴﺚ
M ( x , y , z ) ∈ S ( Ω, R ) ⇔ Ω M = R
ﻟﺪﻳﻨﺎ :
=R
= R2
2
2
) ( x − x0 ) + ( y − y0 ) + ( z − z0
2
⇔
2
) ( x − x0 ) + ( y − y0 ) + ( z − z0
2
⇔
2
) S ( Ω, R
وهﺬﻩ اﻟﻤﻌﺎدﻟﺔ هﻲ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ
أﻣﺜﻠــﺔ :
(1
0
)Ω ( 3, 0,1
R=2
,
S1 : ( x − 3) + ( y ) + ( z − 1) = 4
(2
.R
2
2
)Ω (1, 2, −3
,
2
R =1
S 2 : ( x − 1) + ( y − 2 ) + ( z + 3) = 1
2
2
2
2
)
⎞1
⎛
(3
x + 2 + ( y − 1) + ⎜ z − ⎟ = 2
⎠2
⎝
⎞1
⎛
S1هﻲ ﻓﻠﻜﺔ ﻣﺮآﺰهﺎ ⎟ Ω ⎜ − 2,1,وﺷﻌﺎﻋﻬﺎ . R = 2
⎠2
⎝
R =1
و
) Ω ( 0, 0, 0
(4
2
2
S4 : x 2 + y 2 + z 2 = 1
(2ﻣﻌﺎدﻟﺔ ﻓﻠﻜﺔ ﻣﻌﺮﻓﺔ ﺑﺄﺣﺪ أﻗﻄﺎرهﺎ.
ﻟﺘﻜﻦ Aو Bﻧﻘﻄﺘﻴﻦ ﻣﺨﺘﻠﻔﺘﻴﻦ ﻣﻦ اﻟﻔﻀﺎء . ξ
ﺗﻮﺟﺪ ﻓﻠﻜﺔ وﺣﻴﺪة Sأﺣﺪ أﻗﻄﺎرهﺎ ] . [ AB
ﻟﺘﻜﻦ Sﻓﻠﻜﺔ أﺣﺪ أﻗﻄﺎرهﺎ هﻮ ] . [ AB
ﻟﺘﻜﻦ :
M ∈ S ⇔ AM ⋅ BM = 0
) B ( xB , y B , z B
و
) A ( xA , y A , z A
و:
) . M ( x, y , z
و Sﻓﻠﻜﺔ أﺣﺪ أﻗﻄﺎرهﺎ هﻮ ] . [ AB
(
M ∈ S ⇔ AM ⋅ BM = 0
ﻟﺪﻳﻨﺎ :
⇔ ( x − x A )( x − xB ) + ( y − y A )( y − yB ) + ( z − z A )( z − z B ) = 0
وهﺬﻩ اﻟﻤﻌﺎدﻟﺔ هﻲ ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ Sاﻟﺘﻲ أﺣﺪ أﻗﻄﺎرهﺎ هﻮ ] . [ AB
ﻣﻼﺣﻈﺔ :
إذا آﺎن ] [ ABﻗﻄﺮ ﻟﻠﻔﻠﻜﺔ Sﻓﺈن ﻣﻨﺘﺼﻒ ] [ ABهﻮ ﻣﺮآﺰهﺎ وﺷﻌﺎﻋﻬﺎ هﻮ :
(3
دراﺳﺔ اﻟﻤﻌﺎدﻟﺔ ( E ) : x + y + z − 2ax − 2by − 2cz + d = 0 :
2
2
2
ﻟﺪﻳﻨﺎ :
( E ) ⇔ ( x − a ) + ( y − b ) + ( z − c ) = a 2 + b2 + c2 − d
اﻟﺤﺎﻟﺔ: 1
a 2 + b2 + c2 − d ≺ 0
∅= S
اﻟﺤﺎﻟﺔ: 2
a + b2 + c2 − d = 0
اﻟﺤﺎﻟﺔ: 3
0
a 2 + b2 + c2 − d
ﻧﻀﻊ :
R = a +b +c −d
2
}) S = {Ω ( a, b, c
2
2
2
ﺣﻴﺚ
2
R
0
R = a +b +c −d
) S = S ( Ω ( a , b, c ) , R
2
2
2
ﻣﺜـﺎل :
+ y + z − 2y + z − 3 = 0
2
ﻃـ: 1
ﻟﺪﻳﻨﺎ :
2
2
(E): x
a=0
b =1
1
2
d = −3
1
17
إذن :
= a 2 + b2 + c2 − d = 1 + + 3
0
4
4
17
⎞1
⎛
= .R
⎟ Ω ⎜ 0,1, −وﺷﻌﺎﻋﻬﺎ
إذن S :ﻓﻠﻜﺔ ﻣﺮآﺰهﺎ :
2
⎠2
⎝
c=−
ﻃـ: 2
2
ﻟﺪﻳﻨﺎ :
⎞1
1
⎛
+ ⎜ z + ⎟ = 1+ + 3
⎠2
4
⎝
2
)( E ) : x 2 + ( y − 1
2
1 ⎞ 17
⎛
= ⎟ x + ( y − 1) + ⎜ z +
⇔
⎠2
4
⎝
⎛ ⎛
⎞ −1 ⎞ 17
S = S ⎜⎜ Ω ⎜ 0,1, ⎟ ,
⎟
⎠⎟ 2 ⎠ 2
⎝ ⎝
2
2
AB
2
.
-IIﺗﻘﺎﻃﻊ ﻓﻠﻜﺔ وﻣﺴﺘﻮى :
اﻟﻮﺿﻊ اﻟﻨﺴﺒﻲ ﻟﻤﺴﺘﻮى وﻓﻠﻜﺔ :
ﻟﻴﻜﻦ Pﻣﺴﺘﻮى و Sﻓﻠﻜﺔ ﻣﺮآﺰهﺎ Ωوﺷﻌﺎﻋﻬﺎ . R
ﻟﺪراﺳﺔ اﻟﻮﺿﻊ اﻟﻨﺴﺒﻲ ﻟﻠﻤﺴﺘﻮى Pواﻟﻔﻠﻜﺔ ، S
) ) d = d ( Ω, ( P
ﻧﺤﺴﺐ اﻟﻤﺴﺎﻓﺔ dﺑﻴﻦ ) ( Pو . Ω
) ) d ( Ω, ( P
اﻟﺤﺎﻟﺔ: 1
R
اﻟﺤﺎﻟﺔ: 2
d ( Ω, ( P ) ) = R
∅= S∩P
} ( S ) ∩ ( P ) = {H
ﺑﺤﻴﺚ Hهﻲ اﻟﻤﺴﻘﻂ اﻟﻌﻤﻮدي ﻟﻠﻨﻘﻄﺔ Ωﻋﻠﻰ ) . ( P
ﻓﻲ هﺬﻩ اﻟﺤﺎﻟﺔ ﻧﻘﻮل ان اﻟﻤﺴﺘﻮى ) ( Pﻣﻤﺎس ﻟﻠﻔﻠﻜﺔ ) . ( S
d ( Ω, ( P ) ) ≺ R
اﻟﺤﺎﻟﺔ: 3
ﻓﻲ هﺬﻩ اﻟﺤﺎﻟﺔ ﺗﻘﺎﻃﻊ ) ( Sو ) ( Pهﻮ داﺋﺮة
ﻣﺮآﺰهﺎ . H
) ﺣﻴﺚ Hهﻮ اﻟﻤﺴﻘﻂ اﻟﻌﻤﻮدي ﻟﻠﻨﻘﻄﺔ Ωﻋﻠﻰ ) .( ( Pوﺷﻌﺎﻋﻬﺎ rﺣﻴﺚ :
. r = R2 − d 2
d = d ( Ω, ( P ) ) = ΩH
ﻋﻠﻤﺎ أن :
ﻣﺜـﺎل :
( P ) : 2x − y + z +1 = 0
و:
}S {Ω, 2
ﺣﻴﺚ :
)Ω (1, −1,1
2 +1+1+1
ﻟﺪﻳﻨﺎ :
4 +1+1
= ) ) d ( Ω, ( P
5 5 6
=
2
6
6
∅= S∩P
إذن :
ﻣﻌﺎدﻟﺔ اﻟﻤﺴﺘﻮى اﻟﻤﻤﺎس ﻟﻠﻔﻠﻜﺔ ﻓﻲ ﻧﻘﻄﺔ ﻣﻌﻴﻨﺔ.
ﻟﺘﻜﻦ Aﻧﻘﻄﺔ ﻣﻦ اﻟﻔﻠﻜﺔ Sذات اﻟﻤﺮآﺰ Ωواﻟﺸﻌﺎع . R
وﻟﻴﻜﻦ ) ( Pاﻟﻤﺴﺘﻮى اﻟﻤﻤﺎس ﻟﻠﻔﻠﻜﺔ Sﻓﻲ . A
=
ﻟﺪﻳﻨﺎ :
ﻣﻼﺣﻈﺔ :
ﻣﺜـﺎل :
M ∈ P ⇔ AΩ ⋅ AM = 0
ΩAﻣﻨﻈﻤﻴﺔ ﻋﻠﻰ ) . ( P
) ( Sﻓﻠﻜﺔ ﻣﻌﺎدﻟﺘﻬﺎ :
و:
x + y + z − 2x + 2 y − 2 = 0
2
) A (1,1, 0
ﺣﺪد ﻣﻌﺎدﻟﺔ اﻟﻤﺴﺘﻮى اﻟﻤﻤﺎس ﻟﻠﻔﻠﻜﺔ Sﻓﻲ . A
A∈ S
ﻟﺪﻳﻨﺎ :
إذن :
M ∈ ( P ) ⇔ AΩ ⋅ AM = 0
وﺑﻤﺎأن :
) Ω (1, −,1, 0
) A (1,1, 0
) M ( x, y , z
2
2
) AΩ ( 0, −2, 0
ﻓﺈن :
) AM ( x − 1, y − 1, z
−2 ( y − 1) = 0
إذن :
وﻣﻨﻪ :
y = 1هﻲ ﻣﻌﺎدﻟﺔ اﻟﻤﺴﺘﻮى اﻟﻤﻤﺎس ﻟﻠﻔﻠﻜﺔ Sﻓﻲ اﻟﻨﻘﻄﺔ . A
-IIIﺗﻘﺎﻃﻊ ﻓﻠﻜﺔ وﻣﺴﺘﻘﻴﻢ :
ﻣﺜــﺎل : 1
أدرس ﺗﻘﺎﻃﻊ اﻟﻔﻠﻜﺔ Sواﻟﻤﺴﺘﻘﻴﻢ ) . ( D
ﺣﻴﺚ :
( S ) : x 2 + y 2 + z 2 − 3x + 2 y − 4 z − 5 = 0
) ∈ (t
و:
ﻟﺪراﺳﺔ ﺗﻘﺎﻃﻊ اﻟﻔﻠﻜﺔ ) ( Sواﻟﻤﺴﺘﻘﻴﻢ ) ، ( D
) ∈ (t
ﻧﺤﻞ اﻟﻨﻈﻤﺔ :
⎧x = t
( D ) : ⎪⎨ y = t + 1
⎪z = 2
⎩
⎧x = t
⎪ y = t +1
⎪
⎨
⎪z = 2
⎪⎩ x 2 + y 2 + z 2 − 3 x + 2 y − 4 z − 5 = 0
اﻟﺠــﻮاب :
t + ( t + 1) + 4 − 3t + 2 ( t + 1) − 8 − 5 = 0
2
2
2t 2 + t − 6 = 0
) ∆ = 1 − 4 ( 2 ) × ( −6
ﻟﻨﺤﻞ اﻟﻤﻌﺎدﻟﺔ :
= 49 0
−1 − 7
إذن :
= t1
= −2
4
−1 + 7 3
= t2
=
4
2
وﻣﻨﻪ ﺗﻘﺎﻃﻊ اﻟﻔﻠﻜﺔ Sواﻟﻤﺴﺘﻘﻴﻢ ) ( Dهﻲ اﻟﻨﻘﻄﺘﻴﻦ A ( −2, −1, 2 ) :
⎞ ⎛3 5
⎟. B⎜ , ,2
⎠ ⎝2 2
و:
ﻣﺜــﺎل : 2
أدرس اﻟﻮﺿﻊ اﻟﻨﺴﺒﻲ ﻟﻠﻤﺴﺘﻘﻴﻢ ) ( Dواﻟﻔﻠﻜﺔ ) . ( S
ﺣﻴﺚ :
+ z 2 = −4
) ∈ (t
2
)( S ) : ( x − 1) + ( y + 1
2
⎧x = t +1
( D ) : ⎪⎨ y = −1 + 2t /
⎪z = t
⎩
اﻟﺠــﻮاب :
+ t = −4
2
وﻣﻨﻪ :
2
)(1 + t − 1) + ( −1 + 2t + 1
t 2 + 4t 2 + t 2 = −4
6t 2 = −4
6t 2 + 4 = 0
∆ = −96 ≺ 0
∅ = )(S ) ∩ ( D
2
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