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# 高二数学抛物线的简单几何性质

1 MQ 2 O 3

P Q

P 1 A B R

M

y 2 = 2 px( p > 0)
AB RAB

l1
f (k )

M (?1,0) F l1
l CD

y2 = 4 x
k 1

P1 l2

P2

P

P1 P2
k A B

l1

l2 P f (k )

2

4

y 2 = 2 px
l CD O OA OB

5 O 6 C AB

y 2 = 2 px( p > 0)
N

l1

l2 l2
AN = 3

M

l1

l2
N

N ∈ l1

A B AMN

AM = 7
C

BN = 6

7小

y 2 = 2 px(0 < p < 1)

( x ? 5) 2 + y 2 = 9
x

x

( x ? 6) 2 + y 2 = 27

P
8

PQ

l A(?1 , 0) C B (0 , 8)

C l C

x
l

9

ABCD

AB

l

y = x+4

C D

y2 = x

ABCD

10个

d ×10 4 km
11.

30°

x2 + y 2 = 4x
2

l D

l

A

B

C

AB + CD

1

1

M

Q

MQ//x

p y 2 = 2 px, y = k ( x ? ) 2

P(

p( k 2 + 1 + 1) 2 p(1 + k 2 + 1) p( k 2 + 1 ? 1) 2 p(1 ? k 2 + 1) , ), , ) Q ( k k 2k 2 2k 2 2k (1 + k 2 + 1) ? 2(1 ? k 2 + 1) OP x. y= x, y = k ( k 2 + 1 + 1) 2
p 2
M

x=?

yM =

p(1 ? k 2 + 1) = yQ k y 2 = 2 px

y1
M ( x3 , y3 ) y2 = ? p2 y1
y 2 x1 = 1 p
2

y2
x OP

y1 y2 = ? p 2 ky 2 ? 2 py ? kp 2 = 0
y y= 1x x1

P( x1 , y1 )

Q ( x2 , y 2 )
y1 y2 = ? p 2

y 2 = 2 px

p y = k(x ? ) 2

p x=? 2

? py1 y3 = 2 x1

P( x1 , y1 )

py1 p p2 y3 = = (? py1 ) ? 2 = ? = y2 2 x1 y1 y1
MQ QF P x 2 RAB 1

y = yo y=

p y M (? , y0 ), Q( 0 , y0 ) 2 2p p ) 2
x ,y P

2

MO

y=? yo

2 y0 p

2 py 0
yo ? p
2 2

(x ?

AB
y 2 = 2 px
x

AB

y = x?

p 2
RAB

y 2 ? 2 py ? p 2 = 0

∴ AB = 2 y1 ? y2 = 2 ? ( y1 + y2 ) 2 ? 4 y1 y2 = 4 p
R l l AB

y = x+b b=

y ? 2 py + 2 pb = 0
2

? = 4 p 2 ? 8 pb = 0,
RAB 3

p p R( , p) 2 2 1 AB ? h = 2 p 2 2

AB

h=

2 p 2

l2
1

P

F

l2
y = k ( x + 1)

k

f (k ) k 2 x 2 + ( 2 k 2 ? 4) x + k 2 = 0

f (k )

l1

y 2 = 4x

4 ? 2k 2 2? k2 P P2 ( x2 , y2 ) P( x, y ) x1 + x2 = ,x = 1 ( x1 , y1 ) k2 k2 2 2?k2 2 2? k2 y = k ( x + 1 ) y = P ( , ) x= k2 k k2 k
2

y 2 = 4x
2

F (1,0)

l2
k ≠0

l2

f = (k )
4

{k ? 1 < k < 0
l l l

0 < k < 1}

2 k k k2 = = 2 2?k 1? k 2 ? 1 k2 ? = ( 2 k 2 ? 4) 2 ? 4 k 4 > 0
k ∈ (?1,0)
C D

f (k ) =

1 1? k 2
0 < k <1

?1 < k < 0

f (k )

CD CD

l
2 (2 pt 2 ,2 pt 2 )

A B 0 l C D

(2 pt12 ,2 pt1 )

kCD =
CD CD

1 t1 + t 2

p y = ?(t1 + t 2 ) ? ( x ? ) 2 2 2 ( p (t1 + t 2 ), p (t1 + t 2 )) l p 2 p (t1 + t 2 ) = ?(t1 + t 2 )[ p (t12 + t 2 )? ] 2 1 2 p(t1 + t 2 ) ≠ 0 t12 + t 2 + =0 2
l CD

l

1 2 p (t1 + t 2 )(t12 + t 2 + )=0 2
l F l CD

CF = DF

C ( x1 , y1 ), D( x2 , y2 ) x1 = x2 , y1 = ? y2
5 CD l N

x=? y=0

p 2
l

( x0 , y0 )
2 y12 = 2 px1 , y2 = 2 px2

x0 y0

( x y) A( x1 , y1 ), B ( x2 , y2 ), N ( x0 , y0 ),
Q OA ⊥ OB

∴ x1 x2 =

2 y12 ? y2 4 p2

∴ kOA ? kOB = ?1 ∴ y1 y2 = ?4 p 2
AB
2

x1 x2 + y1 y2 = 0 ∴

2 y12 y2 + y1 y2 = 0 4 p2

Q y1 y2 ≠ 0
N

y ? y0 = ?

x0 ( x ? x0 ), y0

x0 ≠ 0

y y ? ( x + y0 ) ∴x = 0 ? x0
∴ x0 ≠ 0 ∴ y1 y2 =

y 2 = 2 px,

2 2 x0 y 2 + 2 py 0 y ? 2 p ( x0 + y0 )=0

2 2 ? 2 p ( x0 + y0 ) x0 2 2 ? 2 p ( x0 + y0 ) x0 2 2 x0 + y0 ? 2 px0 = 0( x0 ≠ 0)

? 4 p2 =
x y N

x0 y0
OA OB
2 2

x 2 + y 2 ? 2 px = 0( x ≠ 0) A(2 pt 2 ,2 pt )
2 2 4 2

OA

( x ? pt ) + ( y ? pt ) = p (t + t ) B (2 pt1 ,2 pt1 )
2

x + y ? 2 pt ? 2 pty = 0
2 2 2

OA OB

t1t = ?1 ? t1 = ?

t 2 ( x 2 + y 2 ) ? 2 px + 2 pty = 0

1 1 OB ? t1 t t (1 + t 2 )( x 2 + y 2 ? 2 px) = 0 ∴ x 2 + y 2 ? 2 px = 0( x ≠ 0)
3

6

C

N C

l2

l1

x

MN C N

O

l2

A B

C

xA

xB

A B

y 2 = 2 px( p > 0)( x A ≤ x ≤ xB , y > 0), p p MN = p, M (? ,0), N ( ,0) Q AM = 17 , AN = 3 2 2 ? (x + ? ? A ? ?( x ? A ? ?
p 2 ) + 2 px A = 17 2 p 2 ) + 2 px A = 9 2

?p = 4 ? ? xA = 1

?p = 2 ? ?xA = 2

AMN

p > xA 2
C

p=4 p = 6?2 = 4 2

xA = 1
C

B

∴ xB = BN ?

y 2 = 8 x(1 ≤ x ≤ 4, y > 0).
P 维

7

P 维

PQ

A( x A , y A ), B ( xB , y B ), C ( xC , yC ), D ( xD , y D ), P ( x1 , y1 ), Q ( x2 , y2 ) 个

2 2 ? ?( x ? 5) + y = 9 ? 2 ? ? y = 2 px

x 2 ? 2(5 ? p ) x + 16 = 0

∴ x1 =

x A + xB = 5? P个 2

y1 =

2p y A + yB = ( x A + xB ) 2 2 2p = x A + xB + 2 x A xB 个 2 2p = 2(5 ? p ) + 8 2 = 9 p ? p2

2 2 ? ?( x ? 6) + y = 27 ? 2 ? ? y = 2 px

x 2 ? 2( 6 ? p ) x + 9 = 0

∴ x2 =

xC + xD = 6? p个 2

y2 =

2p yC + y D = ( xC + xD ) 2 2
S ?ABQ = S ?APQ + S ?BPQ =

y1

y2 = 9 p ? p 2 2P 2

x1 ? x2 = 1, y1 ? y2 = 0 ∴ PQ = 1 个
2P 10 ? 2 p ? 8 = 2 p (1 ? p ) 个

1 PQ ? y A ? y B = 2

x A ? xB =

4

Q0 < p < 1
8
'

p=

1 2
C

S ?ABQ
A

y 2 = 2 px ( p > 0) l ' A(?1 , 0) B(0 , 8) l A ( x1 , y1 ) x1 ? 1 x ? y1 ? y2 + 8 ? k 2 ?1 = k ? , =k? 2 , x = , ? ? 1 ? 2 2 2 ?2 ? ? 2 k +1 ? y ? ? y ?8 ? y = ? 2k ; ? 2 ? 1 ? k = ?1, ? k = ?1, 1 2 ? ? ? x + 1 x k + 1 ? ? 1 ? 2
A' B'

l ∠B Ox = α C

1 2 B

l

y = kx (k ≠ 0) B ' ( x2 , y 2 )
16k ? x2 = 2 , ? ? k +1 ? 2 ? y = 8(k ? 1) . ? 2 k 2 +1 ?

? 4k 2 k 2 ?1 = 2 p ? , ? 2 2 k 2 +1 ? (k + 1) ? 2 2 ? 64(k ? 1) = 2 p ? 16k . 2 2 ? k 2 +1 ? (k + 1)
p p=
2 5 5

k 2 ? k ?1 = 0 l A B l

1± 5 1+ 5 p>0 k >0 k= 2 2 1+ 5 4 5 y= x C y2 = x 2 5 A' ( x1 , y1 ) B ' ( x2 , y2 ) k=
OB ' = OB = 8

k=

1+ 5 2

∠B 'Ox = α

OA' = OA = 1 ∠B 'OA' = 90°

x2 = 8 cos α

y2 = 8 sin α

∠BOA = 90°

x1 > 0
' '

x2 > 0

α

x1 = cos(α ? 90°) = sin α A' (sin α , ? cos α )

y1 = sin(α ? 90°) = ? cos α B ' (8 cos α , 8 sin α )

2 ? ?cos α = 2 p sin α , A B ? 2 ? ?64 sin α = 16 p cos α . 1 5 2 5 2 5 8 sin 3 α = cos 3 α tan α = sin α = cos α = p= C 2 5 5 5 4 5 90° ? α α y2 = x l ∠B 'OB l α+ = + 45° 5 2 2 α sin(α + 90°) cos α 1+ 5 1+ 5 k = tan( + 45°) = = = l y= x 2 1 + cos(α + 90°) 1 ? sin α 2 2

(1) (2) 9

AB

y = x+4

AB // CD

CD

y = x+b

C ( x1 , y1 )

D ( x2 , y 2 )

? y2 = x ? ?y = x + b

x

y2 ? y + b = 0

5

y1 + y2 = 1

y1 y2 = b

CD = 1 +

1 y1 ? y2 ( k2

k = 1) ABCD

CD = 2 ? ( y1 + y2 ) 2 ? 4 y1 y2 = 2(1 ? 4b)

2(1 ? 4b) = 4?b 2

CD

AB

b = ?6

2 b + 8b + 12 = 0 b = ?6 b = ?2 2 ABCD S = CD = 2(1 + 24) = 50
2

CD

CD =

4?b

b = ?2

ABCD

S = CD = 2(1 + 8) = 18

2

ABCD

18

50

10

y 2 = 2 px
P ( x0 , y0 ) PF

p>0

F(

p , 0) 2

y=

3 p (x ? ) 3 2 (7 ± 4 3 ) p 2
(4 ± 2 3 ) p = d p= 2± 3 d 2 P

? y 2 = 2 px, ? ? 3 p ( x ? ), ?y = 3 2 ?
PF =

x=

(7 ± 4 3 ) p 2

x0 =

2 3 p 2 3 (7 ± 4 3 ) p p | x0 ? |= | ? |= (4 ± 2 3 ) p 3 2 3 2 2
p 2± 3 = d 2 4

2+ 3 d × 10 4 km 4
(2)

2? 3 d ×10 4 km 4 p , 0) 2

F
(1)

P ( x0 , y0 ) PF =
11

y 2 = 2 px x0 = 0

F(

x=?

p 2

p p + x0 ≥ ( x0 ≥ 0) 2 2

PF

AB + CD
x2 + y 2 = 4x F ( 2 , 0) F ( 2 , 0)
2

( x ? 2) 2 + y 2 = 4

y 2 = 8x

AB + CD = AD ? BC AB + CD = AD ? 4 A( x1 , y1 )

BC
A

BC = 4
D

D ( x2 , y 2 ) AD = AF + FD l y = 2( x ? 2)
y x2 ? 6x + 4 = 0

? y 2 = 8, ? ? y = 2( x ? 2). AD

x1 + x2 = 6

AD = 6 + 4 = 10 AB + CD

AB + CD = 10 ? 4 = 6 AD ? BC = AD ? 4

AB

CD
6

### 20081202高二数学(2.4.2 抛物线的简单几何性质 第一课....ppt

20081202高二数学(2.4.2 抛物线的简单几何性质 第一课时)_数学_高中教育_教育专区。2.4 2.4.2 抛物线 抛物线的简单几何性质 第一课时 问题提出 ? 1 ? 5730 p?...