椭圆x^2+y^2/4=1/4及直线y=x+m
⑴当直线和椭圆有共公点,求实数m的取值范围.
y=x+m代入椭圆方程中得:4x^2+x^2+2mx+m^2-1=0
5x^2+2mx+m^2-1=0
判别式=4m^2-20(m^2-1)>=0
16m^2<=20
m^2<=5/4
-根号5/2<=m<=根号5/2.
⑵求被椭圆截得的最长玄所在的直线方程.
x1+x2=-2m/5,x1x2=(m^2-1)/5
弦长=根号(1+k^2)*|x2-x1|=根号(1+1)*根号[(x1+x2)^2-4x1x2]=根号2*根号(4m^2/25-4(m^2-1)/5)=根号2*根号[(-16m^2+20)/25]
所以,当m=0时,弦长为最长.此时直线方程是y=x.