已知抛物线的顶点是C(0,a)(a>0,a为常数),并经过点(2a,2a),点D(0,2a)已知抛物线的顶点是C(0,a)(a>0,a为常数),并经过点(2a,2a),点D(0,2a)为一定点.(1)求含有常数a的抛物线的解析式;(2)
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![已知抛物线的顶点是C(0,a)(a>0,a为常数),并经过点(2a,2a),点D(0,2a)已知抛物线的顶点是C(0,a)(a>0,a为常数),并经过点(2a,2a),点D(0,2a)为一定点.(1)求含有常数a的抛物线的解析式;(2)](/uploads/image/z/6920781-69-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E9%A1%B6%E7%82%B9%E6%98%AFC%280%2Ca%29%28a%3E0%2Ca%E4%B8%BA%E5%B8%B8%E6%95%B0%29%2C%E5%B9%B6%E7%BB%8F%E8%BF%87%E7%82%B9%282a%2C2a%29%2C%E7%82%B9D%280%2C2a%29%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E9%A1%B6%E7%82%B9%E6%98%AFC%EF%BC%880%2Ca%EF%BC%89%EF%BC%88a%EF%BC%9E0%2Ca%E4%B8%BA%E5%B8%B8%E6%95%B0%EF%BC%89%2C%E5%B9%B6%E7%BB%8F%E8%BF%87%E7%82%B9%EF%BC%882a%2C2a%EF%BC%89%2C%E7%82%B9D%EF%BC%880%2C2a%EF%BC%89%E4%B8%BA%E4%B8%80%E5%AE%9A%E7%82%B9%EF%BC%8E%EF%BC%881%EF%BC%89%E6%B1%82%E5%90%AB%E6%9C%89%E5%B8%B8%E6%95%B0a%E7%9A%84%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%EF%BC%9B%EF%BC%882%EF%BC%89)
已知抛物线的顶点是C(0,a)(a>0,a为常数),并经过点(2a,2a),点D(0,2a)已知抛物线的顶点是C(0,a)(a>0,a为常数),并经过点(2a,2a),点D(0,2a)为一定点.(1)求含有常数a的抛物线的解析式;(2)
已知抛物线的顶点是C(0,a)(a>0,a为常数),并经过点(2a,2a),点D(0,2a)
已知抛物线的顶点是C(0,a)(a>0,a为常数),并经过点(2a,2a),点D(0,2a)为一定点.
(1)求含有常数a的抛物线的解析式;
(2)设点P是抛物线上任意一点,过P作PH丄x轴.垂足是H,求证:PD=PH;
(3)设过原点O的直线l与抛物线在笫一象限相交于A、B两点,若DA=2DB.且S△ABD=4 2.求a的值.
已知抛物线的顶点是C(0,a)(a>0,a为常数),并经过点(2a,2a),点D(0,2a)已知抛物线的顶点是C(0,a)(a>0,a为常数),并经过点(2a,2a),点D(0,2a)为一定点.(1)求含有常数a的抛物线的解析式;(2)
(1)设抛物线的解析式为y=kx2+a
∵点D(2a,2a)在抛物线上,
4a2k+a = 2a ∴k =
∴抛物线的解析式为y= x2+a
(2)设抛物线上一点P(x,y),过P作PH⊥x轴,PG⊥y轴,在Rt△GDP中,
由勾股定理得:PD2=DG2+PG2=(y–2a)2+x2 =y2 – 4ay+4a2+x2
∵y= x2+a ∴x2 = 4a ´ (y– a)= 4ay– 4a2 (6分)
∴PD 2= y2– 4ay+4a2 +4ay– 4a2= y2 =PH2
∴PD = PH
(3)过B点BE ⊥ x轴,AF⊥x轴.
由(2)的结论:BE=DB AF=DA
∵DA=2DB ∴AF=2BE ∴AO = 2BO
∴B是OA的中点,
∴C是OD的中点,
连结BC
∴BC= = = BE = DB
过B作BR⊥y轴,
∵BR⊥CD ∴CR=DR,OR= a + = ,
∴B点的纵坐标是,又点B在抛物线上,
∴ = x2+a ∴x2 =2a2
∵x>0 ∴x = a
∴B (a, )
AO = 2OB, ∴S△ABD=S△OBD = 4
所以,´2a´a= 4
∴a2= 4 ∵a>0 ∴a = 2
:(1)设抛物线的解析式为y=kx2+a
∵点D(2a,2a)在抛物线上,
4a2k+a = 2a ∴k =
∴抛物线的解析式为y= x2+a
(2)设抛物线上一点...
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:(1)设抛物线的解析式为y=kx2+a
∵点D(2a,2a)在抛物线上,
4a2k+a = 2a ∴k =
∴抛物线的解析式为y= x2+a
(2)设抛物线上一点P(x,y),过P作PH⊥x轴,PG⊥y轴,在Rt△GDP中,
由勾股定理得:PD2=DG2+PG2=(y–2a)2+x2 =y2 – 4ay+4a2+x2
∵y= x2+a ∴x2 = 4a ´ (y– a)= 4ay– 4a2 (6分)
∴PD 2= y2– 4ay+4a2 +4ay– 4a2= y2 =PH2
∴PD = PH
(3)过B点BE ⊥ x轴,AF⊥x轴.
由(2)的结论:BE=DB AF=DA
∵DA=2DB ∴AF=2BE ∴AO = 2BO
∴B是OA的中点,
∴C是OD的中点,
连结BC
∴BC= = = BE = DB
过B作BR⊥y轴,
∵BR⊥CD ∴CR=DR,OR= a + = ,
∴B点的纵坐标是,又点B在抛物线上,
∴ = x2+a ∴x2 =2a2
∵x>0 ∴x = a
∴B (a, )
AO = 2OB, ∴S△ABD=S△OBD = 4
所以,´2a´a= 4
∴a2= 4 ∵a>0 ∴a = 2
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