Question

求一高二数学题答案 在线等

已知抛物线y²=-x与直线1:y=k(x+1)相交于A,B两点，O为原点坐标 ①当k=1时，求线段AB的长度 ②求证OA垂直OB ③当三角形AOB的面积是根号10时，求直线L的方程

Answer 1

1
y=x+1
(x+1)^2=-x
x^2+3x+1=0
(x+3/2)^2=5/4
x1=(-3-√5)/2 ,y1=(-1-√5)/2
x2=(-3+√5)/2 y2=(-1+√5)/2
AB^2=(x1-x2)^2+(y1-y2)^2=2(2√5)^2=100
AB=10

2
y1/x1=(-1-√5)/(-3-√5)=(1+√5)/(3+√5)=(3-√5)/(√5-1)=-x2/y2
OA垂直OB

3
y=k(x+1)过M(-1,0)
y^2=-x
k^2(x+1)^2=-x
k^2x^2+(2k^2+1)x+k^2=0
x1+x2=(2k^2+1)/k^2
x1x2=1
(x1-x2)^2=(x1+x2)^2-4x1x2=(2k^2+1)^2/k^4-4=(4k^2+1)/k^4
|x1-x2|=√(4k^2+1)/k^2
(y1-y2)=k(x1-x2) |y1-y2|= √(4k^2+1)/|k|
S=|y1-y2|*OM /2=√10
√(4k^2+1)/|2k|=√10 (4k^2+1)/4k^2=10 1/4k^2=9
k^2=1/36 k1=1/6,k2=-1/6
L: y=(1/6)(x+1) 或y=(-1/6)(x+1)