已知圆o的方程为x^2+y^2=9 求过点A(1,2)的圆的弦的中点P的轨迹.设过点A(1,2)的圆的弦所在直线y=k(x-1)+2=kx+(2-k)与圆O交于A(x1,y1),B(x2,y2)两点中点P的坐标(x0,y0)2x0=x1+x22y0=y1+y2x1^2+y1^2=9.1式x2^2+y2^2=9.2式2-1
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![已知圆o的方程为x^2+y^2=9 求过点A(1,2)的圆的弦的中点P的轨迹.设过点A(1,2)的圆的弦所在直线y=k(x-1)+2=kx+(2-k)与圆O交于A(x1,y1),B(x2,y2)两点中点P的坐标(x0,y0)2x0=x1+x22y0=y1+y2x1^2+y1^2=9.1式x2^2+y2^2=9.2式2-1](/uploads/image/z/8708581-37-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86o%E7%9A%84%E6%96%B9%E7%A8%8B%E4%B8%BAx%5E2%2By%5E2%3D9+%E6%B1%82%E8%BF%87%E7%82%B9A%281%2C2%29%E7%9A%84%E5%9C%86%E7%9A%84%E5%BC%A6%E7%9A%84%E4%B8%AD%E7%82%B9P%E7%9A%84%E8%BD%A8%E8%BF%B9.%E8%AE%BE%E8%BF%87%E7%82%B9A%281%2C2%29%E7%9A%84%E5%9C%86%E7%9A%84%E5%BC%A6%E6%89%80%E5%9C%A8%E7%9B%B4%E7%BA%BFy%3Dk%28x-1%29%2B2%3Dkx%2B%282-k%29%E4%B8%8E%E5%9C%86O%E4%BA%A4%E4%BA%8EA%28x1%2Cy1%29%2CB%28x2%2Cy2%29%E4%B8%A4%E7%82%B9%E4%B8%AD%E7%82%B9P%E7%9A%84%E5%9D%90%E6%A0%87%28x0%2Cy0%292x0%3Dx1%2Bx22y0%3Dy1%2By2x1%5E2%2By1%5E2%3D9.1%E5%BC%8Fx2%5E2%2By2%5E2%3D9.2%E5%BC%8F2-1)
已知圆o的方程为x^2+y^2=9 求过点A(1,2)的圆的弦的中点P的轨迹.设过点A(1,2)的圆的弦所在直线y=k(x-1)+2=kx+(2-k)与圆O交于A(x1,y1),B(x2,y2)两点中点P的坐标(x0,y0)2x0=x1+x22y0=y1+y2x1^2+y1^2=9.1式x2^2+y2^2=9.2式2-1
已知圆o的方程为x^2+y^2=9 求过点A(1,2)的圆的弦的中点P的轨迹.
设过点A(1,2)的圆的弦所在直线y=k(x-1)+2=kx+(2-k)
与圆O交于A(x1,y1),B(x2,y2)两点
中点P的坐标(x0,y0)
2x0=x1+x2
2y0=y1+y2
x1^2+y1^2=9.1式
x2^2+y2^2=9.2式
2-1式
(x2+x1)(x2-x1)+(y2-y1)(y1+y2)=0
(x2+x1)(x2-x1)+k(x2-x1)(y1+y2)=0
当x1≠x2时
(x2+x1)+k(y1+y2)=0
2x0+2ky0=0
k=-x0/y0
因为中点也在直线上
直线斜率k=(y0-2)/(x0-1)
所以通过中间参数k,得到等式
-x0/y0=(y0-2)/(x0-1)
即
-x/y=(y-2)/(x-1)
中点P的轨迹
(x-1/2)^2+(y-1)^2=5/4
是个圆
当x1=x2,也满足(x-1/2)^2+(y-1)^2=5/4
所以综上
中点P的轨迹
(x-1/2)^2+(y-1)^2=5/4
为什么要2式-1式?这么做的意义是?(本人学渣……先谢谢各位了)
已知圆o的方程为x^2+y^2=9 求过点A(1,2)的圆的弦的中点P的轨迹.设过点A(1,2)的圆的弦所在直线y=k(x-1)+2=kx+(2-k)与圆O交于A(x1,y1),B(x2,y2)两点中点P的坐标(x0,y0)2x0=x1+x22y0=y1+y2x1^2+y1^2=9.1式x2^2+y2^2=9.2式2-1
如果参加高考,这样解题,就没有希望了,奥林匹克高手告诉你简单的方法啦
高考方法一:
P(x,y)
OP⊥弦AB
k(OP)*k(AB)=-1
(y/x)*(y-2)/(x-1)=-1
(x-0.5)^2+(y-1)^2=1.25
方法二:
P(x,y),弦AB
xA+xB=2x,yA+yB=2y
k(AB)=(yA-yB)/(xA-xB)=(y-2)/(x-1)
(xA)^2+(yA)^2=9.(1)
(xB)^2+(yB)^2=9.(2)
(1)-(2):
(xA+xB)*(xA-xB)+(yA+yB)*(yA-yB)=0
(xA+xB)+(yA+yB)*(yA-yB)/(xA-xB)=0
2x+2y*(y-2)/(x-1)=0
x^2-x+y^2-2y=0
(x-0.5)^2+(y-1)^2=1.25
如果奥林匹克高手这样解都看不明白,那你要改文科了.
设直线斜率为k P(x,y)弦AB A(x1,y1) B(x2,y2)x1^2+y1^2=9x2^2+yy^2=9 相减(x1+x2)(x1-x2)+(y1+y2)(y1-y2)=0x1+x2=2x y1+y2=2y (y1-y2)/(x1-x2)=k所以 2x+2yk=0 x+ky=0k=(y-2)/(x-1)x+y(y-2)&#47...
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设直线斜率为k P(x,y)弦AB A(x1,y1) B(x2,y2)x1^2+y1^2=9x2^2+yy^2=9 相减(x1+x2)(x1-x2)+(y1+y2)(y1-y2)=0x1+x2=2x y1+y2=2y (y1-y2)/(x1-x2)=k所以 2x+2yk=0 x+ky=0k=(y-2)/(x-1)x+y(y-2)/(x-1)=0x^2-x+y^2-2y=0圆的弦的中点P的轨迹为 x^2-x+y^2-2y=0
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