题目内容
如图,抛物线y=﹣x2+3x+4与x轴交于A、B两点,与y轴交于C点,点D在抛物线上且横坐标为3.
(1)求tan∠DBC的值;
(2)点P为抛物线上一点,且∠DBP=45°,求点P的坐标.
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(1)求tan∠DBC的值;
(2)点P为抛物线上一点,且∠DBP=45°,求点P的坐标.
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(1)tan∠DBC=
;
(2)P(﹣
,
).
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(2)P(﹣
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试题分析:(1)连接CD,过点D作DE⊥BC于点E.利用抛物线解析式可以求得点A、B、C、D的坐标,则可得CD//AB,OB=OC,所以∠BCO=∠BCD=∠ABC=45°.由直角三角形的性质、勾股定理和图中相关线段间的关系可得BC=4
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(2)过点P作PF⊥x轴于点F.由∠DBP=45°及∠ABC=45°可得∠PBF=∠DBC,利用(1)中的结果得到:tan∠PBF=
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试题解析:
(1)令y=0,则﹣x2+3x+4=﹣(x+1)(x﹣4)=0,
解得 x1=﹣1,x2=4.
∴A(﹣1,0),B(4,0).
当x=3时,y=﹣32+3×3+4=4,
∴D(3,4).
如图,连接CD,过点D作DE⊥BC于点E.
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∵C(0,4),
∴CD//AB,
∴∠BCD=∠ABC=45°.
在直角△OBC中,∵OC=OB=4,
∴BC=4
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在直角△CDE中,CD=3.
∴CE=ED=
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∴BE=BC﹣DE=
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∴tan∠DBC=
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(2)过点P作PF⊥x轴于点F.
∵∠CBF=∠DBP=45°,
∴∠PBF=∠DBC,
∴tan∠PBF=
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设P(x,﹣x2+3x+4),则
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解得 x1=﹣
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∴P(﹣
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