如图:抛物线与x轴交与A(-1,0)、B(3,0)两点,与y轴交于点C(0,-3),设抛物线的顶点为D(1)求该抛物线的解析式与顶点D的坐标:(2)求证:△BCD的外接圆的圆心落在BD边的中点:(3)若点P是坐标轴
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![如图:抛物线与x轴交与A(-1,0)、B(3,0)两点,与y轴交于点C(0,-3),设抛物线的顶点为D(1)求该抛物线的解析式与顶点D的坐标:(2)求证:△BCD的外接圆的圆心落在BD边的中点:(3)若点P是坐标轴](/uploads/image/z/6876115-43-5.jpg?t=%E5%A6%82%E5%9B%BE%3A%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%B8%8Ex%E8%BD%B4%E4%BA%A4%E4%B8%8EA%28-1%2C0%29%E3%80%81B%283%2C0%29%E4%B8%A4%E7%82%B9%2C%E4%B8%8Ey%E8%BD%B4%E4%BA%A4%E4%BA%8E%E7%82%B9C%280%2C-3%29%2C%E8%AE%BE%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E9%A1%B6%E7%82%B9%E4%B8%BAD%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%A5%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%E4%B8%8E%E9%A1%B6%E7%82%B9D%E7%9A%84%E5%9D%90%E6%A0%87%EF%BC%9A%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E2%96%B3BCD%E7%9A%84%E5%A4%96%E6%8E%A5%E5%9C%86%E7%9A%84%E5%9C%86%E5%BF%83%E8%90%BD%E5%9C%A8BD%E8%BE%B9%E7%9A%84%E4%B8%AD%E7%82%B9%EF%BC%9A%EF%BC%883%EF%BC%89%E8%8B%A5%E7%82%B9P%E6%98%AF%E5%9D%90%E6%A0%87%E8%BD%B4)
如图:抛物线与x轴交与A(-1,0)、B(3,0)两点,与y轴交于点C(0,-3),设抛物线的顶点为D(1)求该抛物线的解析式与顶点D的坐标:(2)求证:△BCD的外接圆的圆心落在BD边的中点:(3)若点P是坐标轴
如图:抛物线与x轴交与A(-1,0)、B(3,0)两点,与y轴交于点C(0,-3),设抛物线的顶点为D(1)求该抛物线的解析式与顶点D的坐标:(2)求证:△BCD的外接圆的圆心落在BD边的中点:(3)若点P是坐标轴上的一动点,能使得以P,A,C为顶点的三角形与△BCD相似?若存在,请写出点P的坐标:若不存在,请说明理由.
如图:抛物线与x轴交与A(-1,0)、B(3,0)两点,与y轴交于点C(0,-3),设抛物线的顶点为D(1)求该抛物线的解析式与顶点D的坐标:(2)求证:△BCD的外接圆的圆心落在BD边的中点:(3)若点P是坐标轴
(1) y = ax^2 + bx + c
代入A,B,C的坐标:
A:a -b + c = 0
B:9a + 3b + c = 0
C:c = 3
a = -1,b = 2,c = 3
y = -x^2 + 2x + 3
(2) y = -x^2 + 2x + 3 = -(x - 1)^2 +4
D(1,4)
抛物线对称轴为x = 1
P是C的以x = 1为对称轴的对称点时,|DC| = |DP|,△PDC是等腰三角形
P的横坐标为2,P(2,3)
(3) BC斜率k1 = (3-0)/(0-3) = -1
CD斜率k2 = (4-3)/(1-0) = 1
k1*k2 = -1
BC与CD垂直,DM与CB平行,DM斜率-1,方程为 y - 4 = -(x -1) (点斜式)
y = -x + 5 = -x^2 + 2x + 3
x^2 -3x + 2 = 0
(x-1)(x-2) = 0
x = 1 (顶点D,舍去)
x = 2
与抛物线交点M(2,3)
(X+1)(X-3)=0
(1,-4)
(1)y=-x^2+4x-3 D(1,-8)
(2)根据抛物线焦点到顶点的关系可知bcds是直角等边三角形