易知直线y=k(x+根号3)过点N(-根号3,0)
而由椭圆方程可知其焦点在x轴上,且a²=4,b²=1,那么:c²=a²-b²=3,即c=根号3
所以可知椭圆的两个焦点就是点M(根号3,0)和点N(-根号3,0)
也就是说直线y=k(x+根号3)过左焦点N
而由椭圆的定义可得:AM+AN=2a=4,BM+BN=2a=4
所以:△ABM周长=AB+AM+BM=AN+BN+AM+BM=4a=8
已知M(根号3,0),椭圆x^2/4+y^2=1与直线y=k(x+根号3)交与A,B,则三角形ABM的周长为多少
已知M(根号3,0),椭圆x^2/4+y^2=1与直线y=k(x+根号3)交与A,B,则三角形ABM的周长为多少
图怎么画,直线过A,B,左焦点?
图怎么画,直线过A,B,左焦点?
数学人气:668 ℃时间:2020-03-24 15:59:50
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