已知与曲线C:x2+y2-2x-2y+1=0相切的直线L分别交x轴,y轴于A(a,0),B(0,b)两点(a>2,b >2),求线段AB中点的轨迹方程
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![已知与曲线C:x2+y2-2x-2y+1=0相切的直线L分别交x轴,y轴于A(a,0),B(0,b)两点(a>2,b >2),求线段AB中点的轨迹方程](/uploads/image/z/3706759-55-9.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%8E%E6%9B%B2%E7%BA%BFC%3Ax2%2By2-2x-2y%2B1%3D0%E7%9B%B8%E5%88%87%E7%9A%84%E7%9B%B4%E7%BA%BFL%E5%88%86%E5%88%AB%E4%BA%A4x%E8%BD%B4%2Cy%E8%BD%B4%E4%BA%8EA%EF%BC%88a%2C0%EF%BC%89%2CB%EF%BC%880%2Cb%EF%BC%89%E4%B8%A4%E7%82%B9%EF%BC%88a%3E2%2Cb+%3E2%EF%BC%89%2C%E6%B1%82%E7%BA%BF%E6%AE%B5AB%E4%B8%AD%E7%82%B9%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B)
已知与曲线C:x2+y2-2x-2y+1=0相切的直线L分别交x轴,y轴于A(a,0),B(0,b)两点(a>2,b >2),求线段AB中点的轨迹方程
已知与曲线C:x2+y2-2x-2y+1=0相切的直线L
分别交x轴,y轴于A(a,0),B(0,b)两点(a>2,b >2),求线段AB中点的轨迹方程
已知与曲线C:x2+y2-2x-2y+1=0相切的直线L分别交x轴,y轴于A(a,0),B(0,b)两点(a>2,b >2),求线段AB中点的轨迹方程
(x-1)^2+(y-1)^2=1
圆心(1,1),半径=1
直线x/a+y/b=1
bx+ay-ab=0
圆心到切线距离=半径
所以|b+a-ab|/√(a^2+b^2)=1
(a+b-ab)^2=a^2b^2
AB中点x=a/2,y=b/2
a=2x,b=2y
代入
(2x+2y-4xy)^2=4x^2+4y^2
x^2+y^2+4x^2y^2+2xy-4x^2y-4xy^2=x^2+y^2
2x^2y^2+xy-2x^2y-2xy^2=0
a,b都不等于0
所以x,y也不等于0
2xy+1-2x-2y=0
其中x=a/2>1,y=b/2>1
AB中点的轨迹方程:y=kx+b
L与曲线C相切
L:y=kx+b
x^2+y^2-2x-2y+1=0
x^2+(kx+b)^2-2x-2(kx+b)+1=0
(1+k^2)x^2+(2kb-2k-2)x+1-2b+b^2=0
(2kb-2k-2)^2-4(1+k^2)*(1-2b+b^2)=0
[(K(B-1)-1]^2-(1+k^2)*(...
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AB中点的轨迹方程:y=kx+b
L与曲线C相切
L:y=kx+b
x^2+y^2-2x-2y+1=0
x^2+(kx+b)^2-2x-2(kx+b)+1=0
(1+k^2)x^2+(2kb-2k-2)x+1-2b+b^2=0
(2kb-2k-2)^2-4(1+k^2)*(1-2b+b^2)=0
[(K(B-1)-1]^2-(1+k^2)*(1-2b+b^2)=0
k^2*(b-1)^2-2k(b-1)+1-(1-2b+b^2)-k^2*(1-2b+b^2)=0
-2k(b-1)+1-(1-2b+b^2)=0
k=(2b-b^2)/(2b-2)
L:y=[(2b-b^2)/(2b-2)]x+b
x=0,y=b>2
y=0,x=(2-2b)/(2-b)=a
线段AB中点:y=b/2>1,b=2y
段AB中点的轨迹方程:
x=a/2=(1-b)/(2-b)=(1-2y)/(2-2y)>1
2xy-2x-2y+1=0,x>1,y>1
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