题目内容
如图是我省某地一座抛物线形拱桥,桥拱在竖直平面内,与水平桥面相交于A,B两点,拱桥最高点C到AB的距离为9m,AB=36m,D,E为拱桥底部的两点,且DE∥AB,点E到直线AB的距离为7m,则DE的长为______m.
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如图所示,建立平面直角坐标系.
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设AB与y轴交于点H,
∵AB=36,
∴AH=BH=18,
由题可知:
OH=7,CH=9,
∴OC=9+7=16,
设该抛物线的解析式为:y=ax2+k,
∵顶点C(0,16),
∴抛物线y=ax2+16,
代入点(18,7)
∴7=18×18a+16,
∴7=324a+16,
∴324a=-9,
∴a=-
,
∴抛物线:y=-
x2+16,
当y=0时,0=-
x2+16,
∴-
x2=-16,
∴x2=16×36=576
∴x=±24,
∴E(24,0),D(-24,0),
∴OE=OD=24,
∴DE=OD+OE=24+24=48,
故答案为48.

设AB与y轴交于点H,
∵AB=36,
∴AH=BH=18,
由题可知:
OH=7,CH=9,
∴OC=9+7=16,
设该抛物线的解析式为:y=ax2+k,
∵顶点C(0,16),
∴抛物线y=ax2+16,
代入点(18,7)
∴7=18×18a+16,
∴7=324a+16,
∴324a=-9,
∴a=-
1 |
36 |
∴抛物线:y=-
1 |
36 |
当y=0时,0=-
1 |
36 |
∴-
1 |
36 |
∴x2=16×36=576
∴x=±24,
∴E(24,0),D(-24,0),
∴OE=OD=24,
∴DE=OD+OE=24+24=48,
故答案为48.
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