Question

已知抛物线y^2=-x与直线y=k（x+1）相交于A.B两点O为原点求证OA垂直OB，当三角形AOB面积为根号10,求k的值

过程如下：首先，有两个焦点k<0
两式联立 k^2x^2+(2k+1)x+k^2=0
x1+x2=2k^2+1/-k
x1+x2=1

垂直嘛，就是证x1x2+y1y2=0
x1x2+y1y2=1+k^2[x1x2+(x1+x2)+1]
=1+k^2(1+2k^2+1/-k+1)
=1+k^2-2k^2-1+k^2=0得证

1/2(根x1^2+y1^2*根下x2^2+yx^2)=2-(x1^2x2+x2^2x1)=40
x1x2(x1+x2)=-38
(2k^2+1)/-k^2=-38
k^2=1/36
k=-1/6
谁看的懂这步（x1^2+y1^2）*（x2^2+yx^2）=40
x1^2x2^2+(x1^2+y2^2+x2^2y1^2)=40
我真的想弄明白，请好心人帮帮忙，谢谢

Answer 1

1/2(根x1^2+y1^2*根下x2^2+yx^2)=10
根号[（X1方+Y1方）（X2方+Y2方）=20
（X1方+Y1方）（X2方+Y2方）=400
（X1方-X1）（X2方-X2）=400
X1（X1-1）X2（X2-1）=400
X1X2（X1X2-X1-X2+1）=400
X1X2=1
X1+X2=(2k^2+1)/-k^2
得(2k^2+1)/k^2=400-2=398
后面就自己解吧