椭圆:3x^2+4y^2=12
直线:y=x
==>7x^2-12=0
设交点为(x1,y1)(x2,y2)
==>x1+x2=0,x1x2=-12/7
==>|AB|^2=(x1-x2)^2+(y1-y2)^2
=2(x1-x2)^2
=2(x1+x2)^2-8x1x2
=96/7
==>|AB|=4sqrt(42)/7