过抛物线 (C:x^2=2py) 上点 (M) 作抛物线 (D:y^2=4x) 的两条切线 (l_1,l_2) ，切点分别为 (P,Q) ，若 ( riangle MPQ) 的重心为 (GBig(1,dfrac32Big)) ，则 (p=underline{qquadqquad}) .
解析：设点 (M(x_0,dfrac{x_0^2}{2p}),P(x_1,y_1),Q(x_2,y_2)) ，切线设为
[y-dfrac{x_0^2}{2p}=k(x-x_0)
]
与抛物线 (y^2=4x) 联立得
[egin{align}y^2-dfrac4ky+dfrac{2x_0^2}{pk}-4x_0=0end{align}
]
由 (Delta=0) 得
[Big(dfrac4kBig)^2-4Big(dfrac{2x_0^2}{pk}-4x_0Big)=0
]
化简得
[2x_0pk^2-x_0^2k+2p=0
]
所以
[k_1+k_2=dfrac{x_0}{2p},k_1k_2=dfrac{1}{x_0}
]
由 (Delta=0) 得方程 ((1)) 的解为 (y=dfrac2k) ，所以 (y_1=dfrac{2}{k_1},y_2=dfrac2{k_2}) . 故
[egin{align}dfrac{x_0^2}{2p}+y_1+y_2=dfrac{x_0^2}{2p}+dfrac{2}{k_1}+dfrac{2}{k_2}=dfrac{x_0^2}{2p}+2Big(dfrac{k_1+k_2}{k_1k_2}Big)=dfrac{3x_0^2}{2p}=dfrac{9}{2}end{align}
]
而
[egin{align}x_0+x_1+x_2&=x_0+Big(dfrac{y_1}{k_1}-dfrac{x_0^2}{2pk_1}+x_0Big)+Big(dfrac{y_2}{k_2}-dfrac{x_0^2}{2pk_2}+x_0Big)\[1ex]&=x_0+Big(dfrac{2}{k_1^2}-dfrac{x_0^2}{2pk_1}+x_0Big)+Big(dfrac{2}{k_2^2}-dfrac{x_0^2}{2pk_2}+x_0Big)\[1ex]&=2Big[dfrac{(k_1+k_2)^2-2k_1k_2}{k_1^2k_2^2}Big]-dfrac{x_0^2}{2p}Big(dfrac{k_1+k_2}{k_1k_2}Big)+3x_0\[1ex]&=dfrac{x_0^4}{4p^2}-x_0=3end{align}
]
由((2),(6)) 解得 (p=dfrac{3}{16}) .