若椭圆MX^2+NX^2=1与直线X+Y-1=0交于A,B两点,过原点和线段AB中点的直线的斜率为跟2/2则M/N的值等于
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![若椭圆MX^2+NX^2=1与直线X+Y-1=0交于A,B两点,过原点和线段AB中点的直线的斜率为跟2/2则M/N的值等于](/uploads/image/z/6236302-22-2.jpg?t=%E8%8B%A5%E6%A4%AD%E5%9C%86MX%5E2%2BNX%5E2%3D1%E4%B8%8E%E7%9B%B4%E7%BA%BFX%2BY-1%3D0%E4%BA%A4%E4%BA%8EA%2CB%E4%B8%A4%E7%82%B9%2C%E8%BF%87%E5%8E%9F%E7%82%B9%E5%92%8C%E7%BA%BF%E6%AE%B5AB%E4%B8%AD%E7%82%B9%E7%9A%84%E7%9B%B4%E7%BA%BF%E7%9A%84%E6%96%9C%E7%8E%87%E4%B8%BA%E8%B7%9F2%2F2%E5%88%99M%2FN%E7%9A%84%E5%80%BC%E7%AD%89%E4%BA%8E)
若椭圆MX^2+NX^2=1与直线X+Y-1=0交于A,B两点,过原点和线段AB中点的直线的斜率为跟2/2则M/N的值等于
若椭圆MX^2+NX^2=1与直线X+Y-1=0交于A,B两点,过原点和线段AB中点的直线的斜率为跟2/2
则M/N的值等于
若椭圆MX^2+NX^2=1与直线X+Y-1=0交于A,B两点,过原点和线段AB中点的直线的斜率为跟2/2则M/N的值等于
点差法:
设A(x1,y1),B(x2,y2),AB中点C(x0,y0),
则2x0=x1+x2,2y0=y1+y2;
OC的斜率k=y0/x0=(y1+y2)/(x1+x2)=√2/2
因为A,B在椭圆上,
所以,mx1²+ny1²=1
mx2²+ny2²=1
点差得:m(x1²-x2²)+n(y1²-y2²)=0
m(x1-x2)(x1+x2)=-n(y1-y2)(y1+y2)
m/n=-(y1-y2)(y1+y2)/(x1-x2)(x1+x2)
m/n=-[(y1-y2)/(x1-x2)]*[(y1+y2)/(x1+x2)]
(y1+y2)/(x1+x2)=√2/2,
(y1-y2)/(x1-x2)是AB的斜率,AB:x+y-1=0,即y=-x+1,斜率为-1;
所以(y1-y2)/(x1-x2)=-1
所以:m/n=√2/2
注:直线与圆锥曲线相交的题型,与弦的中点有关的题目,点差法是很好的处理方法,比韦达定理法要少很多计算,自己可以到文库里搜一些相关的文档学习学习.
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跟2/2
椭圆mx^2+ny^2=1与直线x+y-1=0交于A,B两点,过原点与线段AB中点的直线斜率为√2/2,则n/m=____√2______
联立椭圆与直线方程得到:
mx^2+ny^2-1=0
y=1-x
===> mx^2+n(x-1)^2-1=0
===> mx^2+nx^2-2nx+n-1=0
===> (m+n)x^2-2nx+(...
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椭圆mx^2+ny^2=1与直线x+y-1=0交于A,B两点,过原点与线段AB中点的直线斜率为√2/2,则n/m=____√2______
联立椭圆与直线方程得到:
mx^2+ny^2-1=0
y=1-x
===> mx^2+n(x-1)^2-1=0
===> mx^2+nx^2-2nx+n-1=0
===> (m+n)x^2-2nx+(n-1)=0
===> x1+x2=2n/(m+n)
而,y1+y2=(1-x1)+(1-x2)=2-(x1+x2)=2-[2n/(m+n)]=2m/(m+2n)
所以,AB中点C的坐标为:
Cx=(x1+x2)/2=n/(m+n)
Cy=(y1+y2)/2=m/(m+n)
所以,OC所在直线的斜率Koc=(Cy-0)/(Cx-0)=Cy/Cx
=[m/(m+n)]/[n/(m+n)]
=m/n
=√2/2
所以,n/m=1/(√2/2)=√2
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