Question

数学一道抛物线的题.求详解

Answer 1

F(p/2,0)

l:x = -p/2

显然F为的BC中点,B(b,c)在x轴下方,(b - p/2)/2 = p/2,b = 3p/2

c = -√3p

C(-p/2,√3p)

BC的斜率k = (√3p + √3p)/(-p/2 - 3p/2) = -√3

BC的方程:y = -√3(x - p/2)

代入y² = 2px,12x² - 20px + 3p² = 0

(6x - p)(2x - 3p) = 0

x = p/6 (舍去x = 3p/2,点B)

A(p/6,p/√3)

AF² = (p/6 - p/2)² + ( p/√3 - 0)²

p = 12

A(2,4√3),K(-6,4√3),F(6,0)

S = (1/2)*KA*A的纵坐标

= 16√3