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如图1所示,已知直线y=kx+m与x轴、y轴分别交于点A、C两点,抛物线y=-x...

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热心网友 时间:2024-10-19 15:28

(1)∵抛物线y=-x 2 +bx+c,当x=- 1 2 时,y取最大值 25 4 ,
∴抛物线的解析式是:y=-(x+ 1 2 ) 2 + 25 4 ,即y=-x 2 -x+6;
当x=0时,y=6,即C点坐标是(0,6),
当y=0时,-x 2 -x+6=0,解得:x=2或-3,
即A点坐标是(-3,0),B点坐标是(2,0).
将A(-3,0),C(0,6)代入直线AC的解析式y=kx+m,
得 -3k+m=0 m=6 ,
解得: k=2 m=6 ,
则直线的解析式是:y=2x+6;

(2)过点B作BD⊥AC,D为垂足,
∵S △ABP :S △BPC =1:3,
∴ 1 2 AP?BD 1 2 PC?BD = 1 3 ,
∴AP:PC=1:3,
由勾股定理,得AC= O A 2 +O C 2 =3 5 .
①当点P为线段AC上一点时,过点P作PH⊥x轴,点H为垂足.
∵PH ∥ OC,
∴ PH OC = AP AC = 1 4 ,
∴PH= 3 2 ,
∴ 3 2 =2x+6,
∴x=- 9 4 ,
∴点P(- 9 4 , 3 2 );
② 当点P在CA延长线时,作PG⊥x轴,点G为垂足.
∵AP:PC=1:3,
∴AP:AC=1:2.
∵PG ∥ OC,
∴ PG OC = AP AC = 1 2 ,
∴PG=3,
∴-3=2x+6,x=- 9 2 ,
∴点P(- 9 2 ,-3).
综上所述,点P的坐标为(- 9 4 , 3 2 )或(- 9 2 ,-3).

(3)设直线y= 1 2 x+a与抛物线y=-x 2 -x+6的交点为M(x M ,y M ),N(x N ,y N )(M在N左侧).
则 x 1 = x M y 1 = y N , x 2 = x N y 2 = y N 为方程组 y= 1 2 x+a y=- x 2 -x+6 的解,
由方程组消去y整理,得:x 2 + 3 2 x+a-6=0,
∴x M 、x N 是方程x 2 + 3 2 x+a-6=0的两个根,
∴x M +x N =- 3 2 ,x M ?x N =a-6,
∴y M ?y N =( 1 2 x M +a)( 1 2 x N +a)= 1 4 x M ?x N + a 2 (x M +x N )+a 2 = 1 4 (a-6)- 3 4 a+a 2 .
①存在a的值,使得∠MON已赞过已踩过你对这个回答的评价是?评论收起 ._1uevpeq{zoom:1;background-color:#fff;border:0;margin-bottom:10px;padding:30px 0 20px 42px;position:relative}._1uevpeq.ec-1841{padding:20px 0}._1uevpeq.ec-2246{padding:20px 0 10px}.ec-1841 .y7we4hu{font-size:16px;margin-bottom:-5px}.y7we4hu{color:#7a8f9a;height:25px;line-height:25px;overflow:hidden;position:relative}.y7we4hu h2{margin:0;padding:0}.y7we4hu:after{clear:both;content:" ";display:block;height:0;visibility:hidden}a.tycfu7u{color:#666;float:right;font-size:12px;margin-left:8px;text-decoration:none}.hhhv6ex{color:#666;font-size:13px;line-height:normal;line-height:20px;margin-top:10px}.vnsdjzp{margin-top:15px;position:relative}.vnsdjzp h3{font-weight:400;padding:0}.vnsdjzp a{text-decoration:none}.vnsdjzp em{color:#d81419;font-style:normal}.ec-2246 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热心网友 时间:2024-10-19 15:35

(1)∵抛物线y=-x 2 +bx+c,当x=- 1 2 时,y取最大值 25 4 ,
∴抛物线的解析式是:y=-(x+ 1 2 ) 2 + 25 4 ,即y=-x 2 -x+6;
当x=0时,y=6,即C点坐标是(0,6),
当y=0时,-x 2 -x+6=0,解得:x=2或-3,
即A点坐标是(-3,0),B点坐标是(2,0).
将A(-3,0),C(0,6)代入直线AC的解析式y=kx+m,
得 -3k+m=0 m=6 ,
解得: k=2 m=6 ,
则直线的解析式是:y=2x+6;

(2)过点B作BD⊥AC,D为垂足,
∵S △ABP :S △BPC =1:3,
∴ 1 2 AP?BD 1 2 PC?BD = 1 3 ,
∴AP:PC=1:3,
由勾股定理,得AC= O A 2 +O C 2 =3 5 .
①当点P为线段AC上一点时,过点P作PH⊥x轴,点H为垂足.
∵PH ∥ OC,
∴ PH OC = AP AC = 1 4 ,
∴PH= 3 2 ,
∴ 3 2 =2x+6,
∴x=- 9 4 ,
∴点P(- 9 4 , 3 2 );
② 当点P在CA延长线时,作PG⊥x轴,点G为垂足.
∵AP:PC=1:3,
∴AP:AC=1:2.
∵PG ∥ OC,
∴ PG OC = AP AC = 1 2 ,
∴PG=3,
∴-3=2x+6,x=- 9 2 ,
∴点P(- 9 2 ,-3).
综上所述,点P的坐标为(- 9 4 , 3 2 )或(- 9 2 ,-3).

(3)设直线y= 1 2 x+a与抛物线y=-x 2 -x+6的交点为M(x M ,y M ),N(x N ,y N )(M在N左侧).
则 x 1 = x M y 1 = y N , x 2 = x N y 2 = y N 为方程组 y= 1 2 x+a y=- x 2 -x+6 的解,
由方程组消去y整理,得:x 2 + 3 2 x+a-6=0,
∴x M 、x N 是方程x 2 + 3 2 x+a-6=0的两个根,
∴x M +x N =- 3 2 ,x M ?x N =a-6,
∴y M ?y N =( 1 2 x M +a)( 1 2 x N +a)= 1 4 x M ?x N + a 2 (x M +x N )+a 2 = 1 4 (a-6)- 3 4 a+a 2 .
①存在a的值,使得∠MON已赞过已踩过你对这个回答的评价是?评论收起 ._1uevpeq{zoom:1;background-color:#fff;border:0;margin-bottom:10px;padding:30px 0 20px 42px;position:relative}._1uevpeq.ec-1841{padding:20px 0}._1uevpeq.ec-2246{padding:20px 0 10px}.ec-1841 .y7we4hu{font-size:16px;margin-bottom:-5px}.y7we4hu{color:#7a8f9a;height:25px;line-height:25px;overflow:hidden;position:relative}.y7we4hu h2{margin:0;padding:0}.y7we4hu:after{clear:both;content:" ";display:block;height:0;visibility:hidden}a.tycfu7u{color:#666;float:right;font-size:12px;margin-left:8px;text-decoration:none}.hhhv6ex{color:#666;font-size:13px;line-height:normal;line-height:20px;margin-top:10px}.vnsdjzp{margin-top:15px;position:relative}.vnsdjzp h3{font-weight:400;padding:0}.vnsdjzp a{text-decoration:none}.vnsdjzp em{color:#d81419;font-style:normal}.ec-2246 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