过抛物线y^2=2px焦点f作弦AB,求三角形AOB的面积的最小值

来源:百度知道 编辑:UC知道 时间:2024/05/16 10:23:53
过抛物线y^2=2px焦点f作弦AB,求三角形AOB的面积的最小值
需要证明.过程.

A(x1,y1)B(x2,y2) y2<0
F(p/2,0)
AB的直线方程为
y=k(x-p/2)
x=y/k+p/2
三角形AOB的面积=SAOF+SBOF=(1/2)*(P/2)*|y1|+(1/2)*(p/2)*|y2|
=(p/4)*(y1-y2)
y^2=2px=2p(y/k+p/2)
y^2-2py/k-p^2=0
y1+y2=2p/k
y1y2=-p^2
(y1-y2)^2=(y1+y2)^2-4y1y2
=4p^2/k^2+4p^2
=4p^2(1/k^2+1)
=4p^2(k^2+1)/k^2
当k趋向无穷大时,即AB垂直x轴
(k^2+1)/k^2趋向1
y1-y2=2p
S△AOB=p^2/2

(1).当AB垂直x轴时,AB:x=p/2
A(p/2, p), B(p/2, -p)
AB=2p
S△AOB=p^2/2

(2).当AB不垂直x轴时,AB:y=k(x-p/2),k≠0
代入抛物线:k^2(x^2-px+p^2/4)=2px
k^2x^2-(k^2+2)px+k^2p^2/4=0
所以 x1+x2=(k^2+2)p/k^2, x1*x2=p^2/4
所以 |y1-y2|=√(y1-y2)^2
=√(kx1-kx2)^2
=|k|*√(x1-x2)^2
=|k|*√[(x1+x2)^2-4x1x2]
=|k|*√[(k^2+2)^2p^2/k^4-p^2]
=|2p/k|*√(k^2+1)
所以 S△AOB=|2p/k|*√(k^2+1)*(p/2)*(1/2)
=|p^2/2k|√(k^2+1)
=|p^2/2|√(1+1/k^2)
无最小值。当k趋向无穷大时,S△AOB=p^2/2

综上,△AOB的最小值是p^2/2,此时AB垂直x轴<