设过点M(2,-2p)做抛物线x^2=2py的两条切线,切点连线中点y=6
问题描述：
设过点M(2,-2p)做抛物线x^2=2py的两条切线,切点连线中点y=6
我想问连立后您是怎么得出x1=pk1,x2=pk2
答
y+2p=k(x-2)y=kx-2k-2px^2=2p(kx-2k-2p)x^2-2pkx+4pk+4p^2=0切点连线中点M,yM=6xM=(x1+x2)/2=pk1yM=k1*pk1-2k1-2p=p(k1)^2-2k1-2p=6.(1)(-2pk1)^2-4*1*(4pk1+4p^2)=0p(k1)^2-4k1-4p=0.(2)k1=,p=