Question

急!!!!已知椭圆C:4x^2+y^2=1及直线y=x+m.

.(1)当直线和椭圆有公共点时,求实数m的取值范围.
(2)求被椭圆截得的最长弦所在的直线方程.

RT~过程~谢谢~

Answer 1

代入
5x^2+2mx+m^2-1=0
有公共点则方程有解
所以4m^2-20(m^2-1)>=0
m^2<=5/4
-√5/2<=m<=√5/2

5x^2+2mx+m^2-1=0
x1+x2=-2m/5,x1x2=(m^2-1)/5
(x1-x2)^2=(x1+x2)^2-4x1x2=(-16m^2+20)/25
y=x+m
所以(y1-y2)=[(x1+m)-(x2+m)]^2=(x1-x2)^2
所以弦AB^2=(x1-x2)^2+(y1-y2)^2=2(x1-x2)^2=(-32m^2+40)/25
显然m=0，(-32m^2+40)/25最大
所以y=x