slide1
Download
Skip this Video
Download Presentation
反比例函数与几何面积问题

Loading in 2 Seconds...

play fullscreen
1 / 16

反比例函数与几何面积问题 - PowerPoint PPT Presentation


  • 139 Views
  • Uploaded on

反比例函数与几何面积问题. 问题: 如图,已知点 P ( x , y )是反比例函数 y= ( k>0 )图象上任意一点,过点 P 作 PA⊥x 轴于 A ,且 =6 探究 1 : 求 k 的值。. 1. 1. y. P. O. A. x. 解:∵点 P ( , )在 y= 上 ∴ k= 依题意有 OA=| | , PA=| | ∴ = ·OA·PA= | |= |k|=6

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about '反比例函数与几何面积问题' - pabla


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide3
解:∵点P( , )在y= 上

∴k=

依题意有OA=| | , PA=| |

∴ = ·OA·PA= | |= |k|=6

∵k>0

∴k=12

slide4
小结:反比例函数 y= (k≠0)图象上一点P(x,y)向 x

轴作垂线,垂足为A ,则构成△POA的面积为 |k| ,即当k

一定时, 也为定值。

y

P

O

A

x

slide5
问题:如图,已知点P( , )是反比例函数y=

(k>0)图象上任意一点,过点P作PA⊥x 轴于A,

且 =6

探究2: 延长PO与双曲线交于另一点Q ( , ),过点Q作QB⊥x 轴于B,求证:OP=OQ

y

P

B

O

A

x

Q

slide6
另类做法:

证明:当k=12 时,有y = ①

设过点P直线为 y= x( >0)②

将②代入①有 = x,即 -12=0 ③

依题知 、 是方程③的两根。∴ + =0

而 =OA, = - OB

∴OA-OB=0,即OA=OB

又∵∠OBQ=∠PAO=90°,∠QOB=∠AOP

∴ △OBQ≌△OAP,∴OP=OQ

slide7
小结:若反比例函数 y = (k≠0)与正比例

函数 y = x ( ≠0) 存在两个交点P( , ),Q( , ),则点P与点Q关于原点对称。

y

P

B

O

A

x

Q

slide8
探究3: 延长BQ至点C,过点C作CD⊥y 轴于D,交双曲线于点E,连接QE、BD、QD、BE,

①求△BQD和△BDE的面积。

② 试探索QE与BD是什么位置关系?为什么?

y

P

B

O

A

x

Q

C

D

E

slide9
y

P

B

A

O

x

Q

D

C

E

slide10
y

P

B

A

O

x

Q

C

D

E

slide11
①依题意可知四边形BCDO为矩形,

∴OB=CD

∴ = ·BQ·CD= ·BQ·OB= = 6 连接OE

∴ CD∥OB

∴ = = |K|= ×12=6

slide13
y

B

P

B

M

O

A

x

Q

Q

C

N

D

E

E

D

slide14
②位置关系:QE∥BD

理由:作QM⊥BE于M,EN⊥BD于N,

∴QM∥EN

而 = =6

即 ·BD·QM= ·BD·EN

∴ QM=EN

∴四边形 ENMQ为平行四边形

∴ QE∥BD

slide15
小结 如图,若过点A作AB⊥x 轴于

B,AC⊥y轴于C,分别交双曲线 y=

(k≠0)于点D、E,则有DE∥BC。

y

y

E

A

C

D

D

O

B

O

B

x

x

E

A

C

slide16

ad