已知AB是抛物线y^2=2Px上的点,OA⊥OB,求三角形AOB面积的最小值

为什么要使三角形AOB面积有最小值,则|OA|=|OB|,

为简易起见, 取p > 0, 开口向右.
设OA斜率k, OA方程 y = kx, 代入y^2 = 2px, A(2p/k^2, 2p/k)
OB斜率 -1/k, OB方程 y = -x/k, 代入y^2 = 2px, B(2pk^2, -2pk)
|OA|^2 = (2p/k^2)^2 + (2p/k)^2 = 4p^2(k^2 + 1)/k^4
|OB|^2 = (2pk^2)^2 + (2pk)^2 = 4p^2k^2(k^2 + 1)
|OA|^2 * |OB|^2 = 4p^2k^2*(k^2 + 1) * 4p^2(k^2 + 1)/k^4 = 16p^4*(k^2 + 1)^2/k^2
|OA|*|OB| = 4p^2(k^2+1)/k
S = |OA|*|OB|/2 = 2p^2*(k^2+1)/k = 2p^2*(k + 1/k)
S' = 2p^2*(1 -1/k^2) = 0
k = ±1
二者等价, 只取k = 1:
A(2p, 2p), B(2p, -2p)
|OA|=|OB|

|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