已知圆C:x^2+(y-1)^2=5,直线:mx-y+1-m=0(1)求证:对m∈R,直线l与圆C总有两个不同的交点;(2)设l与圆C交于A,B两点.若AB的绝对值=根号17.求l的倾斜角大小;(3)求弦AB的中点M的轨迹方程.
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![已知圆C:x^2+(y-1)^2=5,直线:mx-y+1-m=0(1)求证:对m∈R,直线l与圆C总有两个不同的交点;(2)设l与圆C交于A,B两点.若AB的绝对值=根号17.求l的倾斜角大小;(3)求弦AB的中点M的轨迹方程.](/uploads/image/z/10239994-10-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86C%3Ax%5E2%2B%28y-1%29%5E2%3D5%2C%E7%9B%B4%E7%BA%BF%3Amx-y%2B1-m%3D0%281%29%E6%B1%82%E8%AF%81%EF%BC%9A%E5%AF%B9m%E2%88%88R%2C%E7%9B%B4%E7%BA%BFl%E4%B8%8E%E5%9C%86C%E6%80%BB%E6%9C%89%E4%B8%A4%E4%B8%AA%E4%B8%8D%E5%90%8C%E7%9A%84%E4%BA%A4%E7%82%B9%EF%BC%9B%EF%BC%882%EF%BC%89%E8%AE%BEl%E4%B8%8E%E5%9C%86C%E4%BA%A4%E4%BA%8EA%2CB%E4%B8%A4%E7%82%B9.%E8%8B%A5AB%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%3D%E6%A0%B9%E5%8F%B717.%E6%B1%82l%E7%9A%84%E5%80%BE%E6%96%9C%E8%A7%92%E5%A4%A7%E5%B0%8F%EF%BC%9B%EF%BC%883%EF%BC%89%E6%B1%82%E5%BC%A6AB%E7%9A%84%E4%B8%AD%E7%82%B9M%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B.)
已知圆C:x^2+(y-1)^2=5,直线:mx-y+1-m=0(1)求证:对m∈R,直线l与圆C总有两个不同的交点;(2)设l与圆C交于A,B两点.若AB的绝对值=根号17.求l的倾斜角大小;(3)求弦AB的中点M的轨迹方程.
已知圆C:x^2+(y-1)^2=5,直线:mx-y+1-m=0
(1)求证:对m∈R,直线l与圆C总有两个不同的交点;
(2)设l与圆C交于A,B两点.若AB的绝对值=根号17.求l的倾斜角大小;
(3)求弦AB的中点M的轨迹方程.
已知圆C:x^2+(y-1)^2=5,直线:mx-y+1-m=0(1)求证:对m∈R,直线l与圆C总有两个不同的交点;(2)设l与圆C交于A,B两点.若AB的绝对值=根号17.求l的倾斜角大小;(3)求弦AB的中点M的轨迹方程.
(1) L:mx-y+1-m=0--->m(x-1)=y-1--->L恒过定点(1,1)
而 1²+(1-1)²<5,即该定点在圆C内
∴对m∈R,直线L与圆C总有两个不同的交点
(2) --->d²(C,L) = r²-(|AB|/2)² = 5-17/4 = 3/4
= (-1+1-m)²/(m²+1)
--->4m²=3m²+1--->m²=1--->m=±1--->L的倾斜角=45°或135°
(3)圆C:x^2+(y-1)^2=5-----------------------1
l:mx-y+1-m=0---------------------------2
联立1,2得
(1+m^2)x^2-2m^2x+m^2-5=0
令X=(x1+x2)/2=m^2/(1+m^2)---------------3
Y=(y1+y2)/2=(m^2-m+1)/(1+m^2)-----------4
X=1-1/(1+m^2)
Y=1-m/(1+m^2)
(X-1+Y-1)^2=(1+m)^2/(1+m^2)
=1+2m/(1+m^2)
=1+2(1-Y)
(X-1+Y-1)^2=1+2(1-Y)
(x-1/2)^2+ (y-1)^2=1/4