《2025年高考总复习首选用卷数学人教版》

4.(2023·浙江校考三模)已知双曲线$\frac{x^{2}}{3}-y^{2}=1$，$F_{1}$，$F_{2}$分别为其左、右焦点，点$P(x_{0},y_{0})$为其右支上一点，在点$P$处作双曲线的切线$l$.
(1)若点$P$的坐标为$(3,\sqrt{2})$，求证：$l$为$\angle F_{1}PF_{2}$的角平分线；
(2)过$F_{1}$，$F_{2}$分别作$l$的平行线$l_{1}$，$l_{2}$，其中$l_{1}$交双曲线于$A$，$B$两点，$l_{2}$交双曲线于$C$，$D$两点，求$S_{\triangle PAB}\cdot S_{\triangle PCD}$的最小值.

答案：
解　
(1)证明:由题意,点$P(x_{0},y_{0})$处的切线l的方程为$\frac{x_{0}}{3}x-y_{0}y = 1$,所以在点$(3,\sqrt{2})$处的切线l的方程为$x-\sqrt{2}y = 1$,切线$l:x-\sqrt{2}y = 1$交x轴于点$Q(1,0)$,则$\frac{\vert QF_{1}\vert}{\vert QF_{2}\vert}=\frac{3}{1}$,$\frac{\vert PF_{1}\vert}{\vert PF_{2}\vert}=\frac{3\sqrt{3}}{\sqrt{3}}=\frac{3}{1}$,即$\frac{\vert QF_{1}\vert}{\vert QF_{2}\vert}=\frac{\vert PF_{1}\vert}{\vert PF_{2}\vert}$,所以l为$\angle F_{1}PF_{2}$的角平分线.
(2)在$P(x_{0},y_{0})$处的切线l的方程为$\frac{x_{0}}{3}x-y_{0}y = 1$,当$y_{0}\neq0$,即点P不为右顶点时,直线l的斜率$k=\frac{x_{0}}{3y_{0}}$,$k^{2}=\frac{x_{0}^{2}}{9y_{0}^{2}}=\frac{3 + 3y_{0}^{2}}{9y_{0}^{2}}=\frac{1}{3}+\frac{1}{3y_{0}^{2}}＞\frac{1}{3}$,联立$\begin{cases}y = k(x + 2)\\x^{2}-3y^{2}-3 = 0\end{cases}$,消去y,得$(1 - 3k^{2})x^{2}-12k^{2}x-12k^{2}-3 = 0$,设$A(x_{1},y_{1})$,$B(x_{2},y_{2})$,则$\begin{cases}x_{1}+x_{2}=\frac{12k^{2}}{1 - 3k^{2}}\\x_{1}x_{2}=\frac{-12k^{2}-3}{1 - 3k^{2}}\\\Delta=12(k^{2}+1)\end{cases}$,所以$\vert AB\vert=\sqrt{1+k^{2}}\vert x_{1}-x_{2}\vert=\frac{2\sqrt{3}(1 + k^{2})}{3k^{2}-1}=\vert CD\vert$,记P到$l_{1}$的距离为$d_{1}$,P到$l_{2}$的距离为$d_{2}$,$F_{1}$到l的距离为$d_{3}$,$F_{2}$到l的距离为$d_{4}$,则$d_{1}d_{2}=d_{3}d_{4}=\frac{\vert\frac{x_{0}}{3}(-2)-1\vert}{\sqrt{y_{0}^{2}+\frac{x_{0}^{2}}{9}}}\cdot\frac{\vert\frac{x_{0}}{3}\cdot2-1\vert}{\sqrt{y_{0}^{2}+\frac{x_{0}^{2}}{9}}}=1$,所以$S_{\triangle PAB}\cdot S_{\triangle PCD}=\frac{1}{2}\vert AB\vert\cdot d_{1}\cdot\frac{1}{2}\vert CD\vert\cdot d_{2}=\frac{1}{4}\vert AB\vert^{2}\cdot1=\frac{3(k^{2}+1)^{2}}{(3k^{2}-1)^{2}}=\frac{1}{3}(1+\frac{4}{3k^{2}-1})^{2}＞\frac{1}{3}$;当$y_{0}=0$,即点P为右顶点时,$\vert AB\vert=\vert CD\vert=\frac{2b^{2}}{a}=\frac{2}{\sqrt{3}}$,$\vert PF_{1}\vert=a + c=2+\sqrt{3}$,$\vert PF_{2}\vert=c - a=2-\sqrt{3}$,所以$S_{\triangle PAB}\cdot S_{\triangle PCD}=\frac{1}{2}\vert AB\vert\cdot\vert PF_{1}\vert\cdot\frac{1}{2}\vert CD\vert\cdot\vert PF_{2}\vert=\frac{1}{2}\times\frac{2}{\sqrt{3}}\times(2+\sqrt{3})\times\frac{1}{2}\times\frac{2}{\sqrt{3}}\times(2-\sqrt{3})=\frac{1}{3}$,综上,$S_{\triangle PAB}\cdot S_{\triangle PCD}$的最小值为$\frac{1}{3}$.

5.(2023·广东广州六中高三模拟)如图，已知椭圆$C_{1}:\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$，抛物线$C_{2}:x^{2}=2py(p\gt0)$，$O$为坐标原点.

(1)若抛物线$C_{2}$的焦点正好为椭圆$C_{1}$的上顶点，求$p$的值；
(2)椭圆$C_{1}$与抛物线$C_{2}$在第一象限的交点为$P(x_{0},y_{0})$，过点$P$但不过原点的直线$l$交椭圆$C_{1}$于点$Q$，交抛物线$C_{2}$于点$M(Q,M$不同于点$P)$，若$M$是线段$PQ$的中点，求$p$的最大值，并求当$p$取最大时直线$l$的斜率.

答案： 解
(1)椭圆$C_{1}:\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$,
所以上顶点为$(0,\sqrt{3})$, 依题意$\frac{p}{2}=\sqrt{3}$,
所以$p = 2\sqrt{3}$.
(2)因为点$P(x_{0},y_{0})$既在椭圆$C_{1}$上,又在抛物线$C_{2}$上,
所以$\frac{x_{0}^{2}}{4}+\frac{y_{0}^{2}}{3}=1$且$x_{0}^{2}=2py_{0}$,
设直线l的方程为$y = kx + m(m\neq0)$,$Q(x_{Q},y_{Q}),M(x_{M},y_{M})$,
联立$\begin{cases}y = kx + m\\\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\end{cases}$,消去y,整理得$(3 + 4k^{2})x^{2}+8kmx + 4m^{2}-12 = 0$,
则$x_{0}+x_{Q}=-\frac{8km}{3 + 4k^{2}}$,
所以$x_{M}=\frac{x_{0}+x_{Q}}{2}=-\frac{4km}{3 + 4k^{2}}$.
联立$\begin{cases}y = kx + m\\x^{2}=2py\end{cases}$,消去y,整理得$x^{2}-2pkx-2pm = 0$,
则$x_{0}x_{M}=-2pm$,所以$x_{0}=\frac{-2pm}{x_{M}}=\frac{p(3 + 4k^{2})}{2k}$,
代入抛物线方程得$y_{0}=\frac{x_{0}^{2}}{2p}=\frac{p(3 + 4k^{2})^{2}}{8k^{2}}$,
再代入椭圆方程得$3\cdot\frac{p^{2}(3 + 4k^{2})^{2}}{4k^{2}}+4\cdot\frac{p^{2}(3 + 4k^{2})^{4}}{(8k^{2})^{2}}=12$, 整理得$\frac{48}{p^{2}}=3(\frac{3}{k}+4k)^{2}+\frac{1}{4}(\frac{3}{k}+4k)^{4}$,
令$t = (\frac{3}{k}+4k)^{2}$,则$\frac{48}{p^{2}}=3t+\frac{1}{4}t^{2}$,
依题意可知$k>0$且$\frac{3}{k}+4k\geqslant2\sqrt{\frac{3}{k}\cdot4k}=4\sqrt{3}$,
当且仅当$\frac{3}{k}=4k$,即$k=\frac{\sqrt{3}}{2}$时取等号,
所以$t = (\frac{3}{k}+4k)^{2}\geqslant48$, 所以$\frac{1}{p^{2}}\geqslant15$,
即$p\leqslant\frac{1}{\sqrt{15}}=\frac{\sqrt{15}}{15}$,
即p的最大值为$\frac{\sqrt{15}}{15}$,此时直线l的斜率为$\frac{\sqrt{3}}{2}$.