中考数学动点与抛物线问题解析
考点二十二 动点与抛物线问题
典型例题:(
如图,已知抛物线y =a (x -1) 2+a ≠0) 经过点A(-2,0) ,抛物线的顶点为D ,过0作射线OM∥AD.过顶点D 平行于x 轴的直线交射线OM 于点C ,B 在x 轴正半轴上,连结BC .
(1)求该抛物线的解析式;
(2)若动点P 从点0出发,以每秒l 个长度单位的速度沿射线OM 运动,设点P 运动的时间为t(s).问:当t 为何值时,四边形DAOP 分别为平行四边形? 直角梯形? 等腰梯形?
(3)若OC=OB,动点P 和动点Q 分别从点O 和点B 同时出发,分别以每秒l 个长度单位和2个长度单位的速度沿OC 和B0运动,当其中一个点停止运动时另一个点也随之停止运动设它们运动的时间为t(s),连接PQ ,当t 为何值时,四边形BCPQ 的面积最小? 并求出最小值及此时PQ 的长.
解:(1)
抛物线y =a (x -1) 2+a ≠0) 经过点A (-2,0) ,
3
∴0=9a +a =-·························································································· 1分
∴
二次函数的解析式为:y =-
3
x +
2
3
x +
3
·················································· 3分
过D 作D N ⊥O B 于N
,则D N =,
(2)
D 为抛物线的顶点∴D (1AN =3,∴AD =
O M ∥A D
=6∴∠DAO =60° ··················································· 4分
①当A D =O P 时,四边形DAOP 是平行四边形
··············································· 5分 ∴OP =6∴t =6(s) ·
②当D P ⊥O M 时,四边形DAOP 是直角梯形
过O 作O H ⊥A D 于H ,A O =2,则A H =1
(如果没求出∠D A O =60°可由R t △O H A ∽R t △D N A 求A H =1)
··························································································· 6分 ∴OP =DH =5t =5(s) ·
③当P D =O A 时,四边形DAOP 是等腰梯形
∴OP =AD -2AH =6-2=4∴t =4(s)
综上所述:当t =6、5、4时,对应四边形分别是平行四边形、直角梯形、等腰梯形. · 7分
△O C B 是等边三角形 (3)由(2)及已知,∠C O B =60°,O C =O B ,
则OB =OC =AD =6,OP =t ,BQ =2t ,∴OQ =6-2t (0
2
过P 作PE ⊥OQ 于E
,则P E =········································································ 8分
·
∴S BC PQ =
12
⨯6⨯12
⨯(6-2t ) ⨯
2
t
3⎫t -+⎪2⎝2⎭2
································································································ 9分 当t =
32
时,S
B C P Q 32
34
··································································10分 ·
34
94
4
∴
此时O Q =3,O P =,
O E =∴Q E =3-
=PE =
∴PQ ==
······················································ 11分 =2
名题精练
1. (2009河南)如图,在平面直角坐标系中,已知矩形ABCD 的三个顶点B (4,0)、C (8,
0)、D (8,8). 抛物线y=ax+bx过A 、C 两点.
(1)直接写出点A 的坐标,并求出抛物线的解析式;
(2)动点P 从点A 出发.沿线段AB 向终点B 运动,同时点Q 从点C 出发,沿线段CD 向终点D 运动.速度均为每秒1个单位长度,运动时间为t 秒. 过点P 作PE ⊥AB 交AC 于点E
①过点E 作EF ⊥AD 于点F ,交抛物线于点G. 当t 为何值时,线段EG 最长?
②连接EQ .在点P 、Q 运动的过程中,判断有几个时刻使得△CEQ 是等腰三角形? 请直接写出相应的t 值.
2
2. 已知二次函数y =ax +bx +c 的图象经过点A (3,0) ,B (2,-3) ,C (0,-3) .
(1)求此函数的解析式及图象的对称轴;
2
(2)点P 从B 点出发以每秒0.1个单位的速度沿线段BC 向C 点运动,点Q 从O 点出发以相同的速度沿线段OA 向A 点运动,其中一个动点到达端点时,另一个也随之停止运动.设运动时间为t 秒.
①当t 为何值时,四边形ABPQ 为等腰梯形;
②设PQ 与对称轴的交点为M ,过M 点作x 轴的平行线交AB 于点N ,设四边形ANPQ 的面积为S ,求面积S 关于时间t 的函数解析式,并指出t 的取值范围;当t 为何值时,S 有 最大值或最小值.
2
3.如图,二次函数y =ax +bx +c (a ≠0)的图象与x 轴交于A 、B 两点,与y 轴相交
、C 两点的坐标分别为A (-3,
于点C .连结A C 、B C ,A 0) 、C (0,且当x =-4和x =2时二次函数的函数值y 相等.
(1)求实数a ,b ,c 的值;
(2)若点M 、N 同时从B 点出发,均以每秒1个单位长度的速度分别沿BA 、BC 边运动,其中一个点到达终点时,另一点也随之停止运动.当运动时间为t 秒时,连结M N ,将△B M N 沿M N 翻折,B 点恰好落在A C 边上的P 处,求t 的值及点P 的坐标;
(3)在(2)的条件下,二次函数图象的对称轴上是否存在点Q ,使得以B ,N ,Q 为项点的三角形与△A B C 相似?如果存在,请求出点Q 的坐标;如果不存在,请说明理由.
考点二十二 答案
1.解.(1)点A 的坐标为(4,8) …………………1 将A (4,8)、C (8,0)两点坐标分别代入y=ax2+bx 8=16a +4b
得
0=64a +8b
解 得a =-12
, b =4
12
∴抛物线的解析式为:y =-x 2+4x …………………3分
PE AP
(2)①在Rt △APE 和Rt △ABC 中,tan ∠PAE =∴PE =
12
=
BC AB
, 即
PE AP
=
48
AP =
12
t .PB=8-t .
12
∴点E的坐标为(4+∴点G 的纵坐标为:-∴EG=- =-∵-181818
t ,8-t ).
12
12
(4+t )2+4(4+
12
t )=-
18
t 2+8. …………………5分
t 2+8-(8-t ) t 2+t .
<0,∴当t =4时,线段EG 最长为2. …………………7分
②共有三个时刻. …………………8分
t 1=
163
, t2=
4013
,t 3
. …………………11分
2. 解:(1)∵二次函数y =ax 2+bx +c 的图象经过点C (0,-3) ,
∴c =-3.
将点A (3,0) ,B (2,-3) 代入
y =ax +bx +⎧0=9a +3b -3,
⎨
⎩-3=4a +2b -3.
2
解得:a =1,b =-2.
∴y =x -2x -3.-------------------2分
2
2
配方得:y =(x -1)-4,所以对称轴为x =1.-------------------3分 (2) 由题意可知:BP = OQ=0.1t . ∵点B ,点C 的纵坐标相等, ∴BC ∥OA .
过点B ,点P 作BD ⊥OA ,PE ⊥OA ,垂足分别为D ,E . 要使四边形ABPQ 为等腰梯形,只需PQ =AB .
即QE =AD =1.
又QE =OE -OQ =(2-0.1t )-0.1t =2-0.2t , ∴2-0.2t =1. 解得t =5.
即t=5秒时,四边形ABPQ 为等腰梯形.-------------------6分 ②设对称轴与BC ,x 轴的交点分别为F ,G . ∵对称轴x =1是线段BC 的垂直平分线, ∴BF =CF =OG =1. 又∵BP =OQ , ∴PF =QG .
又∵∠PMF =∠QMG , ∴△MFP ≌△MGQ . ∴MF =MG .
∴点M 为FG 的中点 -------------------8分 ∴S=S 四边形
=S 四边形由S 四边形
S ∆BPN =
ABPQ
-S ∆BPN ,
ABFG
-S ∆BPN . =1212
(BF +AG ) FG =FG =
340t .
92
ABFG
.
1
∴S=
92
-
2
3
BP ⋅
40
t .-------------------10分
又BC =2,OA =3,
∴点P 运动到点C 时停止运动,需要20秒. ∴0
∴当t =20秒时,面积S 有最小值3.------------------11
⎧9a -3b +c =0,
⎪
3、(1)由题意,得⎨16a -4b +c =4a +2b +c ,
⎪
⎩c =
⎧a =-⎪
3⎪
⎪⎪
解之得⎨b =- ······················································································· 3分 3⎪
⎪c =⎪⎪⎩
33
233
(2)由(1)得y =-x -
2
x +3,当y =0时,x =-3或1.
∴B (1,0),A (-3,0),C (0,3).
∴OA =3,OB =1,OC =3. 易求得AC =23,B C =2,A B =4. ∴△ABC 为Rt △,且∠ACB =90°,∠A =30°,∠B =60°. 又由BM =BN =PN =PM
∴PN ∥AB , ∴
PN AB
=43
CN CB
知四边形P M B N 为菱形, .
,即
t 4
=
2-t 2
∴t =. ·······································································································5分
过P 作PE ⊥AB 于E ,
在Rt △PEM 中,∠PME =∠B =60°,PM =
43
32
43
.
∴PE =PM ⋅sin 60=
ME =
⨯=
2323
3.
PE tan 60
=.
又OM =BM -OB =∴P (-113
故,
.·····························································································7分
(3)由(1)、(2)知抛物线y =-
且∠ACB =90°.
①若∠BQN =90°,
33
x -
2
233
x +3的对称轴为直线x =-1,
∵BN 的中点到对称轴的距离大于1, 而
12BM =
23
∴以BN 为直径的圆不与对称轴相交, ∴∠BQN ≠90°,
即此时不存在符合条件的Q 点. ②若∠BNQ =90°,
当∠NBQ =60°,则Q 、E 重合,此时∠BNQ ≠90°; 当∠NBQ =30°,则Q 、P 重合,此时∠BNQ ≠90°.
即此时不存在符合条件的Q 点.
③若∠QBN =90°时,延长NM 交对称轴于点Q , 此时,Q 为P 关于x 轴的对称点.
∴Q (-1,-23
3)为所求.
分
10