Question

求解答数学题,椭圆方程的 !已知椭圆4x^2＋y^2＝1及直线y＝x＋m（1）当直线与椭圆有交点时,求m的取值范围（2）

Answer 1

把直线方程代入椭圆得

4x^2＋(x+m)^2＝1

5x^2+2mx+m^2-1=0

x1+x2=-2m/5

x1x2=(m^2-1)/5

△≥0

(2m)^2-4*5*(m^2-1)≥0

-√5/2≤m≤√5/2

弦长=√[(x1-x2)^2+(y1-y2)^2]

=√(1+k^2)*√(x1-x2)^2

=√(1+k^2)*√[(x1+x2)^2-4x1x2]

=2√[4m^2/25-4(m^2-1)/5]

=2√[(-16m^2+20)/25]

很明显m=0时有最大值4√5/5