直线x－y＋1＝0被椭圆x²÷16＋y²÷4＝1截得的弦长为多少,

x－y＋1＝0
y=x+1代入x²÷16＋y²÷4＝1
x²÷16＋(x+1)²÷4＝1
x²＋4(x+1)²＝16
5x²+8x-12=0
根据韦达定理：x1+x2=-8/5,x1x2=-2/5
(x2-x1)² = (x2+x1)² -4x1x2 = (8/5)²+412/5 = 304/25
(y2-y1)² = (x2+1-x1-1)² = (x2-x1)² = 304/25
截得的弦长 = √{(x2-x1)²+(y2-y1)²} = √{608/5) = 4√38/5