已知椭圆c：4x∧2+y∧2＝1及直线l：y＝x+m,m属于R

(1)
将直线方程代入椭圆方程：
4x² + (x+m)² = 1
5x² + 2mx + m² -1 = 0 (i)
x = [-m±√(5-4m²)]/5
设二交点为A(a, p), B(b, q):
则p = a + m; q = b + m
二者相减,得 q - p = b - a
|AB| = √[(b -a)² + (q-p)²] = √[2(b -a)²] =(√ 2)|b -a| = (√2)|b -a|
= (√2)| [-m+√(5-4m²)]/5 - [-m - √(5-4m²)]/5|
= (2√2)√(5-4m²)/5 = (2√2)/5
√(5-4m²) = 1
m = ±1
直线方程： y = x ± 1
(2) 从（1）可知A([-m+√(5-4m²)]/5, [4m+√(5-4m²)]/5))
B([-m-√(5-4m²)]/5, [4m-√(5-4m²)]/5)
OA的斜率u = [4m+√(5-4m²)]/[-m+√(5-4m²)]
OB的斜率v = [4m-√(5-4m²)]/[-m-√(5-4m²)]
uv = [16m² -(5-4m²)]/[m² - (5-4m²)]
= (20m² -5)/(5m² -5) = -1
m² = 2/5
m = ±(√10)/5
y = x ±(√10)/5
(3)从（2）可知,AB的中点M(-m/5, 4m/5)
令x = -m/5, 则m = -5x
y = 4m/5 = 4(-5x)/5 = -4x
直线l被椭圆c截得的弦的中点轨迹: y = -4x
(i)中的判别式=4（5-4m²) = 0时,二者相切,此时m² = 5/4
m = ±(√5)/2
x = -m/5
两个切点的横坐标分别为x1 = -(√5)/10和x2 = (√5)/10
弦中点的轨迹为y = -4x (-(√5)/10 < x