已知椭圆x^2/9+y^2/4=1与直线f(x)=p=2x-3y-12只有一个交点
则x=(p+3y+12)/2,[(p+3y+12)/2]^2/9+y^2/4=1
y^2/2+(p+12)y/6+(p+12)^2/36-1=0
y^2+(p+12)y/3+(p+12)^2/18-2=0
[y+(p+12)/6]^2=[(p+12)/6]^2+2-(p+12)^2/18=2-(p+12)^2/36
只有一个交点,2-(p+12)^2/36=0
(p+12)^2/36=2,p+12=+-6√2
p=6√2-12,或p=-6√2-12