椭圆ax2+by2=1与直线x+y-1=0相交于A,B两点,C是AB的中点,若AB=2倍根号2,OC的斜率为根号2/2,
2个回答

直线OC:y=x√2/2

直线AB:x+y-1=0

上式联解得到C点坐标

x+(√2/2)x-1=0

xC=2/(2+√2),yC=√2/(2+√2)

C[2/(2+√2),√2/(2+√2)]

xA+xB=2xC=4/(2+√2),yA+yB=2yC=2√2/(2+√2)

由直线x+y-1=0

得y=1-x,带入椭圆方程

ax²+by²=1

ax²+b(1-x)²=1

(a+b)x²-2bx+b-1=0

xA+xB=2b/(a+b)=4/(2+√2),b=(√2)a

xA*xB=(b-1)/(a+b)=(√2a-1)/(a+√2a)

(xA+xB)²=8/(3+2√2)

(xA-xB)²=(yA-yBV)²=(xA+xB)²-4xA×xB=8/(3+2√2)-4(√2a-1)/(a+√2a)

(xA-xB)²+(yA-yB)²=AB²

2[8/(3+2√2)-4(√2a-1)/(a+√2a)]=(2√2)²=8

2/(3+2√2)-(√2a-1)/(a+√2a)=1

2/(3+2√2)-1=(√2a-1)/(a+√2a)

(-1-2√2)/(3+2√2)=(√2a-1)/(a+√2a)

a=1/3

b=(√2)a=√2/3

椭圆方程为x²/3+√2y²/3=1