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(1)A(-1,0) ,B(2,3)
(2)△ABP最大面积s=
![](https://img.yulucn.com/upload/7/d4/7d4a8bde06adbc5e17d2ee927c3c09d5_thumb.jpg)
; P(
![](https://img.yulucn.com/upload/9/fe/9fe6a25e8c1ae77770a1de0d3fff60a7_thumb.jpg)
,-
![](https://img.yulucn.com/upload/5/4b/54b2f24a273d43719d27ed700d926e05_thumb.jpg)
)
(3)存在;k=
试题分析:(1)将两个解析式联立组成方程组,解方程组即得
要想△ABP的面积最大,则要在要求的抛物线上找到一个点P,使点P到直线AB的距离最大,这时过点P且与AB平行的直线与抛物线只有一个交点,利用根的判别式可确定平移后所得直线的解析式,进而可得点的坐标,求出面积
设圆心为E,连接EQ,直线与x轴交点为H,与y轴交点为F;由已知可得直线与两坐标轴交点的坐标,从而可得直线与坐标轴交点到原点的距离;由圆的切线及相似的知识可得出EQ、QH的长,
再由勾股定理可得要求的值
试题解析:(1)A(-1,0) ,B(2,3)
(2)平移直线AB得到直线L,当L与抛物线只有一个交点时,△ABP面积最大[如图12-1(1)]
设直线L解析式为:
![](https://img.yulucn.com/upload/9/39/9396cd48d338e65afdaaa702cc19dbe4_thumb.jpg)
,
根据
![](https://img.yulucn.com/upload/c/db/cdb862158f82c46a7d7171763f1fc786_thumb.jpg)
,得
判别式△
![](https://img.yulucn.com/upload/5/78/5782ac45c4ff781809c4616221bcba4e_thumb.jpg)
,解得,
代入原方程中,得
![](https://img.yulucn.com/upload/c/e9/ce90b79f68243d27c76a83f297547833_thumb.jpg)
;解得,
![](https://img.yulucn.com/upload/8/95/895a2771eca751f0a0e538ad1ff6eba7_thumb.jpg)
,
∴P(
![](https://img.yulucn.com/upload/9/fe/9fe6a25e8c1ae77770a1de0d3fff60a7_thumb.jpg)
,
![](https://img.yulucn.com/upload/4/49/449db86d204868c947ffb855a0a035d5_thumb.jpg)
)
易求,AB交
![](https://img.yulucn.com/upload/4/0b/40bf1a8704f4e69377484dd6c328cf71_thumb.jpg)
轴于M(0,1),直线L交轴
![](https://img.yulucn.com/upload/4/0b/40bf1a8704f4e69377484dd6c328cf71_thumb.jpg)
于G(0,
![](https://img.yulucn.com/upload/c/66/c661f0825f2617d5387542e8814c6905_thumb.jpg)
)
过M作MN⊥直线L于N,∵OM=1,OA=1,∴∠AMO=45°
∵∠AMN=90,∴∠NMO=45°
在RT△MNE中,∠NMO=45°,MG=
![](https://img.yulucn.com/upload/e/65/e65f366d689a6a402107ebda44ed6244_thumb.jpg)
,[如图12-1(2)]
∴ MN=
![](https://img.yulucn.com/upload/8/d7/8d7bc874f9460a44f6dc5f1ab9e2c84a_thumb.jpg)
,MN即为△ABP的高
由两点间距离公式,求得:AB=
故△ABP最大面积
(3)设在直线
![](https://img.yulucn.com/upload/b/c4/bc4a4b155ff14972bcc3c0bcf138516a_thumb.jpg)
上存在唯一一点Q使得∠OQC=90°
则点Q为以OC的中点E为圆心,OC为直径形成的圆E与直线
![](https://img.yulucn.com/upload/b/c4/bc4a4b155ff14972bcc3c0bcf138516a_thumb.jpg)
相切时的切点,[如图12-2(1)]
由解析式可知:C(
![](https://img.yulucn.com/upload/7/75/775b8aaf879088f4a54261dd46402b44_thumb.jpg)
,0),OC=
![](https://img.yulucn.com/upload/b/86/b8626c641b09bf679a8cc05c9b91e5d6_thumb.jpg)
2 ,则圆E的半径:OE=CE=
![](https://img.yulucn.com/upload/7/07/707440d2a088879c6d785dc133d37dcc_thumb.jpg)
=QE
设直线
![](https://img.yulucn.com/upload/b/c4/bc4a4b155ff14972bcc3c0bcf138516a_thumb.jpg)
与
![](https://img.yulucn.com/upload/b/86/b8626c641b09bf679a8cc05c9b91e5d6_thumb.jpg)
0 、
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