抛物线Y方=12X截直线Y=2x+1所得的弦长等于?
问题描述：
抛物线Y方=12X截直线Y=2x+1所得的弦长等于?
答
代入
4x^2+4x+1=12x
4x^2-8x+1=0
x1+x2=2,x1x2=1/4
所以(x1-x2)^2=(x1+x2)^2-4x1x2=4-1=3
(y1-y2)^2=[(2x1+1)-(2x2+1)]^2=(2x1-2x2)^2=4(x1-x2)^2=12
所以弦长=根号[(x1-x2)^2+(y1-y2)^2]=根号(3+12)=根号15
答
交点坐标可设为 (x1,y1),(x2,y2)弦长公式L = √((x1-x2)^2 + (y1-y2)^2)= √5*((x1+x2)^2-4x1*x2)将y = 2x+1代入前面的 y^2 = 12x得 4*x^2 - 8*x + 1 = 0;x1+x2 = 2x1*x2 = 1/4代入L得,L = √15...