过原点的两条互相垂直的直线分别交抛物线y*y=4p(x+p)(p＞0）于A,B和C,D四点，则/AB/+/CD/最小值等于？

Question

那位高手帮帮忙,,,谢谢啊!!
过原点的两条互相垂直的直线分别交抛物线y*y=4p(x+p)(p＞0）于A,B和C,D四点，则/AB/+/CD/最小值等于？需要答案最后一步和最后结果即可,,,谢谢!

Answer 1

y^2=4p(x+p)
AB斜率k,y=kx
k^2x^2=4p(x+p)
k^2x^2-4px-4p^2=0
Ax+Bx=4p/k^2
AxBx=-4p^2/k^2
(Ax-Bx)^2=[(4p)^2+4p^2k^2]/k^4=4p^2(4p^2+k^2)/k^4
|AB|^2=(k^2+1)(Ax-Bx)^2=(k^2+1)4p^2(4p^2+k^2)/k^4
|AB|=(2p/k^2)√[(k^2+1)(4p^2+k^2)]
CD斜率-1/k，y=-x/k
x^2/k^2=4p(x+p)
x^2/k^2-4px-4p^2=0
Cx+Dx=4pk^2
CxDx=-4p^2k^2
|CD|^2=(1/k^2+1)4p^2k^2(4p^2k^2+1)
|CD|=2p*√(4p^2k^2+1)(k^2+1)

|AB|+|CD|=(2p/k^2)√[(4p^2k^2+1)(k^2+1)]+2p√[(4p^2k^2+1)(k^2+1)]
|AB|>0,|CD|>0
|AB|+|CD|≥2√|AB||CD|
|AB|=|CD|时，|AB|+|CD|最小=2AB
k=1,|AB|+|CD|=4p√(8p^2+2)