抛物线双切线问题（一）

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 \(\triangle MPQ\) 的重心为 \(G\Big(1,\dfrac32\Big)\) ，则 \(p=\underline{\qquad\qquad}\) .

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

\[\begin{align}y^2-\dfrac4ky+\dfrac{2x_0^2}{pk}-4x_0=0\end{align} \]

由 \(\Delta=0\) 得

\[\Big(\dfrac4k\Big)^2-4\Big(\dfrac{2x_0^2}{pk}-4x_0\Big)=0 \]

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

\[\begin{align}\dfrac{x_0^2}{2p}+y_1+y_2=\dfrac{x_0^2}{2p}+\dfrac{2}{k_1}+\dfrac{2}{k_2}=\dfrac{x_0^2}{2p}+2\Big(\dfrac{k_1+k_2}{k_1k_2}\Big)=\dfrac{3x_0^2}{2p}=\dfrac{9}{2}\end{align} \]

而

\[\begin{align}x_0+x_1+x_2&=x_0+\Big(\dfrac{y_1}{k_1}-\dfrac{x_0^2}{2pk_1}+x_0\Big)+\Big(\dfrac{y_2}{k_2}-\dfrac{x_0^2}{2pk_2}+x_0\Big)\\[1ex]&=x_0+\Big(\dfrac{2}{k_1^2}-\dfrac{x_0^2}{2pk_1}+x_0\Big)+\Big(\dfrac{2}{k_2^2}-\dfrac{x_0^2}{2pk_2}+x_0\Big)\\[1ex]&=2\Big[\dfrac{(k_1+k_2)^2-2k_1k_2}{k_1^2k_2^2}\Big]-\dfrac{x_0^2}{2p}\Big(\dfrac{k_1+k_2}{k_1k_2}\Big)+3x_0\\[1ex]&=\dfrac{x_0^4}{4p^2}-x_0=3\end{align} \]

由\((2),(6)\) 解得 \(p=\dfrac{3}{16}\) .