y＝kx+1与椭圆x²/2+y²＝1交于M,N.且绝对值MN＝3分之4倍的根号2,求直线l方程

y=kx+1x^2/2+y^2=1x^2+2*(kx+1)^2=2(1+2k^2)x^2+4kx=0xM+xN=-4k/(1+2k^2),xM*xN=0(xM-xN)^2=(xM+xN)^2-4xM*xN=16k^2/(1+2k^2)^2(yM-yN)^2=k^2*(xM-xN)^2(xM-xN)^2+(yM-yN)^2=MN^2(1+k^2)*16k^2/(1+2k^2)^2=(4√2/3)^...