Question

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

Answer 1

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|