Question

已知直线l过点A（-3/2p,p)且与抛物线y^2=2px（p>0）只有一个公共点,求直线l的方程.

Answer 1

线l过点A(-3p/2,p)且与抛物线y2=2px

Y=KX+(2P+3PK)/2

[KX+(2P+3PK)/2]^2=2PX

判别=0

(3k^2+2k-2)^2=k^2(2+3k)^2

3k^2+2k-2=-k(2+3k)

k=1/3,k=-1

y=x/3+3p,y=-x-p