gmx931024
幼苗
共回答了16个问题采纳率:93.8% 举报
(1)y=
![](https://img.yulucn.com/upload/a/40/a40022f0bd5dee3e87faa0e156ed82a8_thumb.jpg)
x
2 -2x;(2)1.8;(3)(
![](https://img.yulucn.com/upload/4/1b/41b18f00dc0e0e938b95ee60783d6409_thumb.jpg)
,
![](https://img.yulucn.com/upload/7/0f/70f32c42970712f851f82ba143e1a016_thumb.jpg)
)
试题分析:(1)由抛物线y=ax
2 +bx经过点A(4,0)与点(-2,6)即可根据待定系数法求解;
(2)过点O作OF⊥AD,连接AC交OB于点E,由垂径定理得AC⊥OB.根据切线的性质可得AC⊥AD,即可证得四边形OFAE是矩形,由tan∠AOB=
![](https://img.yulucn.com/upload/4/b9/4b940ba018a6b23bf81071530a649915_thumb.jpg)
可得sin∠AOB=
![](https://img.yulucn.com/upload/d/a7/da7128e411c10251cb068a72e0ac3c53_thumb.jpg)
,即可求得AE、OD的长,当PQ⊥AD时,OP=t,DQ=2t.则在Rt△ODF中,OD=3,OF=AE=2.4,DF=DQ-FQ=DQ-OP=2t-t=t,再根据勾股定理求解;
(3)设直线l平行于OB,且与抛物线有唯一交点R(相切),此时△ROB中OB边上的高最大,所以此时△ROB面积最大,由tan∠AOB=
![](https://img.yulucn.com/upload/4/b9/4b940ba018a6b23bf81071530a649915_thumb.jpg)
可得直线OB的解析式为y=
![](https://img.yulucn.com/upload/4/b9/4b940ba018a6b23bf81071530a649915_thumb.jpg)
x,由直线l平行于OB,可设直线l解析式为y=
![](https://img.yulucn.com/upload/4/b9/4b940ba018a6b23bf81071530a649915_thumb.jpg)
x+b.点R既在直线l上,又在抛物线上,可得
![](https://img.yulucn.com/upload/a/40/a40022f0bd5dee3e87faa0e156ed82a8_thumb.jpg)
x
2 -2x=
![](https://img.yulucn.com/upload/4/b9/4b940ba018a6b23bf81071530a649915_thumb.jpg)
x+b,再根据直线l与抛物线有唯一交点R(相切),可得方程2x
2 -11x-4b=0有两个相等的实数根,即可得到判别式△=0,从而可以求得结果.
(1)∵抛物线y=ax
2 +bx经过点A(4,0)与点(-2,6),
∴
![](https://img.yulucn.com/upload/4/2f/42f3637647a8a7ded5fa8c6695927b98_thumb.jpg)
,解得a=
![](https://img.yulucn.com/upload/a/40/a40022f0bd5dee3e87faa0e156ed82a8_thumb.jpg)
,b=-2
∴抛物线的解析式为:y=
![](https://img.yulucn.com/upload/a/40/a40022f0bd5dee3e87faa0e156ed82a8_thumb.jpg)
x
2 -2x;
(2)过点O作OF⊥AD,连接AC交OB于点E,由垂径定理得AC⊥OB.
∵AD为切线,
∴AC⊥AD,
∴AD∥OB.
∴四边形OFAE是矩形,
∵tan∠AOB=
∴sin∠AOB=
![](https://img.yulucn.com/upload/d/a7/da7128e411c10251cb068a72e0ac3c53_thumb.jpg)
,
∴AE=OA·sin∠AOB=4×
![](https://img.yulucn.com/upload/d/a7/da7128e411c10251cb068a72e0ac3c53_thumb.jpg)
=2.4,
OD=OA·tan∠OAD=OA·tan∠AOB=4×
![](https://img.yulucn.com/upload/4/b9/4b940ba018a6b23bf81071530a649915_thumb.jpg)
=3.
当PQ⊥AD时,OP=t,DQ=2t.
则在Rt△ODF中,OD=3,OF=AE=2.4,DF=DQ-FQ=DQ-OP=2t-t=t,
由勾股定理得:DF=
![](https://img.yulucn.com/upload/c/b3/cb3c543d4331b4d1b085e142dd84a477_thumb.jpg)
,
∴t=1.8秒;
(3)设直线l平行于OB,且与抛物线有唯一交点R(相切),
此时△ROB中OB边上的高最大,所以此时△ROB面积最大.
∵tan∠AOB=
∴直线OB的解析式为y=
![](https://img.yulucn.com/upload/4/b9/4b940ba018a6b23bf81071530a649915_thumb.jpg)
x,
由直线l平行于OB,可设直线l解析式为y=
![](https://img.yulucn.com/upload/4/b9/4b940ba018a6b23bf81071530a649915_thumb.jpg)
x+b.
∵点R既在直线l上,又在抛物线上,
∴
![](https://img.yulucn.com/upload/a/40/a40022f0bd5dee3e87faa0e156ed82a8_thumb.jpg)
x
2 -2x=
![](https://img.yulucn.com/upload/4/b9/4b940ba018a6b23bf81071530a649915_thumb.jpg)
x+b,化简得:2x
2 -11x-4b=0.
∵直线l与抛物线有唯一交点R(相切),
∴方程2x
2 -11x-4b=0有两个相等的实数根
∴判别式△=0,即11
2 +32b=0,解得b=
![](https://img.yulucn.com/upload/9/5e/95e21550496e20a439efca2944ba3bf0_thumb.jpg)
,
此时原方程的解为x=
![](https://img.yulucn.com/upload/4/1b/41b18f00dc0e0e938b95ee60783d6409_thumb.jpg)
,即x
R =
![](https://img.yulucn.com/upload/4/1b/41b18f00dc0e0e938b95ee60783d6409_thumb.jpg)
,
而y
R =
![](https://img.yulucn.com/upload/a/40/a40022f0bd5dee3e87faa0e156ed82a8_thumb.jpg)
x
R 2 -2x
R =
∴点R的坐标为R(
![](https://img.yulucn.com/upload/4/1b/41b18f00dc0e0e938b95ee60783d6409_thumb.jpg)
,
![](https://img.yulucn.com/upload/7/0f/70f32c42970712f851f82ba143e1a016_thumb.jpg)
).
点评:此类问题是初中数学的重点和难点,在中考中极为常见,一般以压轴题形式出现,难度较大.
1年前
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