已知过函数y^2=4x的焦点的直线与图像的两个交点构成的等边三角形的另一个点在函数的准线上,求三角形的边

Question

已知过函数y^2=4x的焦点的直线与图像的两个交点构成的等边三角形的另一个点在函数的准线上,求三角形的边

Answer 1

抛物线 y^2=4x,p = 2,焦点(1,0),准线 x=-1
设两交点是 A(x1,y1),B(x2,y2)
若直线垂直x轴,和准线上的 点C所成的等腰三角形 底边AB= p+p =4,高=p =2,不够成等边三角形
所以设 直线是
y-0 = k(x-1)
代入y^2=4x,x1,x2满足方程
k^2(x-1)^2 = 4x
(x-1)^2 = 4/k^2
x^2 -(2+ 4/k^2)x +1 =0
(x1+x2) = 2+ 4/k^2
A,B 到准线的距离是到焦点的距离
（x1- (-p/2)) + (x2 - (-p/2))
= (x1+x2) +p
= 2+ 4/k^2 +2
= 4 + 4/k^2
所以 AB的长度= 4 + 4/k^2
设AB中点是M(x3,y3)
M满足
x3 = 1/2 (x1+x2) = 1+2/k^2
y3 = k( 1+2/k^2 -1) = 2/k
过AB中点作垂线交准线于C(-1,y4)
垂线方程是
y - 2/k = -1/k *(x- (1+2/k^2))
y4 - 2/k = -1/k *(-1 - (1+2/k^2))
y4= 4/k+ 2/k^3
CM 应= AB sin60
CM^2 = (-1 -x3)^2 + (y4-y3)^2 = (2+2/k^2) + (2/k+ 2/k^3)^2 = (1+1/k^2)(2+2/k^2)
(AB sin60)^2 = (4 + 4/k^2)^2 * 3/4= 3 *(2+2/k^2)
所以(1+1/k^2) = 3
1/k^2 = 2
代回AB
AB= 4 + 4/k^2 = 4 + 4 * 2 = 12