Question

直线与圆的方程直线l:x-ky+2倍根号2=0与圆C：x2+y2=4交于A,B两点,O为坐标原点,三角形ABC的面积为S

Answer 1

1.x=ky-2√2代入圆的方程,得(1+k^2)y^2-4√2ky+4=0

x=ky-2√2过M(-2√2,0),不妨假定S(AOB)=|S(AOM)-S(BOM)|=1/2*OM*|y1-y2|＝√2|y1-y2|,y1,y2分别为A,B纵坐标.

|y1-y2|=√[32k^2-16(1+k^2)]/(1+k^2)=4√(k^2-1)/(1+k^2)

S(k)=4√[2(k^2-1)]/(1+k^2),定义域(-∞,1)∪(1,+∞)

2.令k^2=t,t>1,则S(k)/4√2=√(k^2-1)/(1+k^2)=√(t-1)/(1+t)

令p=√(t-1),则p>0,t=p^2+1,S(k)/4√2=p/(p^2+2)=1/(p+2/p)