如图,抛物线y=a(x-1)²+4与x轴交于AB两点,与y轴交于C点 D是抛物线的顶点,如图,抛物线y=a(x-1)²+4与x轴交于AB两点,与y轴交于C点 D是抛物线的顶点,CD=√2,在抛物线上共有三个点到直线BC的
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![如图,抛物线y=a(x-1)²+4与x轴交于AB两点,与y轴交于C点 D是抛物线的顶点,如图,抛物线y=a(x-1)²+4与x轴交于AB两点,与y轴交于C点 D是抛物线的顶点,CD=√2,在抛物线上共有三个点到直线BC的](/uploads/image/z/5290825-49-5.jpg?t=%E5%A6%82%E5%9B%BE%2C%E6%8A%9B%E7%89%A9%E7%BA%BFy%3Da%EF%BC%88x-1%EF%BC%89%26%23178%3B%2B4%E4%B8%8Ex%E8%BD%B4%E4%BA%A4%E4%BA%8EAB%E4%B8%A4%E7%82%B9%2C%E4%B8%8Ey%E8%BD%B4%E4%BA%A4%E4%BA%8EC%E7%82%B9+D%E6%98%AF%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E9%A1%B6%E7%82%B9%2C%E5%A6%82%E5%9B%BE%2C%E6%8A%9B%E7%89%A9%E7%BA%BFy%3Da%EF%BC%88x-1%EF%BC%89%26%23178%3B%2B4%E4%B8%8Ex%E8%BD%B4%E4%BA%A4%E4%BA%8EAB%E4%B8%A4%E7%82%B9%2C%E4%B8%8Ey%E8%BD%B4%E4%BA%A4%E4%BA%8EC%E7%82%B9+D%E6%98%AF%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E9%A1%B6%E7%82%B9%2CCD%3D%E2%88%9A2%2C%E5%9C%A8%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%B8%8A%E5%85%B1%E6%9C%89%E4%B8%89%E4%B8%AA%E7%82%B9%E5%88%B0%E7%9B%B4%E7%BA%BFBC%E7%9A%84)
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如图,抛物线y=a(x-1)²+4与x轴交于AB两点,与y轴交于C点 D是抛物线的顶点,如图,抛物线y=a(x-1)²+4与x轴交于AB两点,与y轴交于C点 D是抛物线的顶点,CD=√2,在抛物线上共有三个点到直线BC的
如图,抛物线y=a(x-1)²+4与x轴交于AB两点,与y轴交于C点 D是抛物线的顶点,
如图,抛物线y=a(x-1)²+4与x轴交于AB两点,与y轴交于C点 D是抛物线的顶点,CD=√2,在抛物线上共有三个点到直线BC的距离为m,求m的值.
如图,抛物线y=a(x-1)²+4与x轴交于AB两点,与y轴交于C点 D是抛物线的顶点,如图,抛物线y=a(x-1)²+4与x轴交于AB两点,与y轴交于C点 D是抛物线的顶点,CD=√2,在抛物线上共有三个点到直线BC的
答:
抛物线y=a(x-1)²+4,开口向下a<0
点C(0,a+4),点D(1,4)
CD=√(1+a²)=√2
解得:a=-1(a=1不符合舍去)
所以:y=-(x-1)²+4
与x轴交点A(-1,0),B(3,0)
点C(0,3),则BC直线为y=-x+3
到BC直线上的点有3个,则BC下侧有
1个,BC上侧有1个
设切线为y=-x+d联立抛物线方程:
y=-x²+2x+3=-x+d
x²-3x+d-3=0
判别式=(-3)²-4(d-3)=0
解得:d=21/4
解得:x=3/2
切线为y=-x+3/2与BC直线y=-x+3的距离为:
(3/2) /√2=3√2/4
所以:m=3√2/4