求抛物线y^2=4X截直线y=2X-1所得的线段|AB|的长度

分类: 作业答案 • 2022-07-10 12:31:21

问题描述：

答

y^2=4x
y=2x-1
y=2 * y^2/4-1
y^2-2y-2=0
AB=√(1+1/k^2)*√[(y1+y2)^2-4y1*y2]
=√5/4*√(2^2+4*2)=√15

答

y^2=4X
y=2X-1 X1=1+√3/2 X2=1-√3/2 A(1+√3/2,1+√3 ) B(1-√3/2,1-√3)
Y1=1+√3 Y2=1-√3
|AB|^2=AB^2= (1+√3/2-1+√3/2)^2+(1+√3-1+√3)^2=15
|AB|=√15

答

把直线代入抛物线得
(2x-1)^2=4x
4x^2-8x+1=0
x1+x2=2,x1x2=1/4
|AB|=√(1+k^2)*√(x1-x2)^2
=√(1+k^2)*√[(x1+x2)^2-4x1x2]
=√5*√3
＝√15

答

y1 =4x1①；y2 =4x2②，①-②得：y1+y2=4/k ③①+②得4(x1+x④又AB=2p(1+1/k ),即2r=2p(1+1/k )=4(1+1/k ) ⑤联立④