《2025年创新设计高考总复习数学浙江专版》

角度 3 一般弦
例 4 (2025·杭州调研)已知顶点在原点，关于$y$轴对称的抛物线与直线$x - 2y = 1$交于$P$，$Q$两点，若$\vert PQ\vert=\sqrt{15}$，则抛物线的方程为 (C)

A.$x^{2}=-4y$
B.$x^{2}=12y$
C.$x^{2}=-4y$或$x^{2}=12y$
D.以上都不是

答案： 例4 C[设抛物线的方程为$x^{2} = 2ay$，则抛物线与直线$x - 2y = 1$联立消去$y$，得$x^{2} - ax + a = 0$，所以$x_{1} + x_{2} = a$，$x_{1}x_{2} = a$，则$\left|x_{1} - x_{2}\right| = \sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \sqrt{a^{2} - 4a}$，所以$\left|PQ\right| = \sqrt{1 + k^{2}}\left|x_{1} - x_{2}\right|=\sqrt{1 + \frac{1}{4}} \cdot \sqrt{a^{2} - 4a} = \sqrt{15}$，所以$a^{2} - 4a - 12 = 0$，解得$a = -2$或$a = 6$，所以$x^{2} = -4y$或$x^{2} = 12y$.]

(1)已知斜率为$2$的直线经过椭圆$\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$的右焦点$F$，且与椭圆相交于$A$，$B$两点，则弦$AB$的长为$\frac{5\sqrt{5}}{3}$.

答案：
(1)$\frac{5\sqrt{5}}{3}$ [
(1)由题意，知椭圆的右焦点$F$的坐标为$(1,0)$，直线$AB$的方程为$y = 2(x - 1)$. 由$\begin{cases}y = 2(x - 1)\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1\end{cases}$消去$y$，得$3x^{2} - 5x = 0$， 设$A(x_{1},y_{1})$，$B(x_{2},y_{2})$，则$x_{1} + x_{2} = \frac{5}{3}$，$x_{1}x_{2} = 0$， 则$\left|AB\right| = \sqrt{(1 + k^{2})[(x_{1} + x_{2})^{2} - 4x_{1}x_{2}]}=\sqrt{(1 + 2^{2}) × [(\frac{5}{3})^{2} - 4 × 0]} = \frac{5\sqrt{5}}{3}$.]

(2)已知抛物线$C:y^{2}=2px(p＞0)$的焦点到准线的距离为$1$，若抛物线$C$上存在关于直线$l:x - y - 2 = 0$对称的不同的两点$P$和$Q$，则线段$PQ$的中点坐标为(1,-1).

答案：
(2)$(1,-1)$ [
(2)
∵焦点到准线的距离为$p$，则$p = 1$，
∴$y^{2} = 2x$.设点$P(x_{1},y_{1})$，$Q(x_{2},y_{2})$， 则$\begin{cases}y_{1}^{2} = 2x_{1}\\y_{2}^{2} = 2x_{2}\end{cases}$，则$(y_{1} - y_{2})(y_{1} + y_{2}) = 2(x_{1} - x_{2})$，
∴$k_{PQ} = \frac{2}{y_{1} + y_{2}}$ 又
∵$P$，$Q$关于直线$l$对称，
∴$k_{PQ} = -1$，即$y_{1} + y_{2} = -2$，
∴$PQ$中点的纵坐标为$\frac{y_{1} + y_{2}}{2} = -1$， 又
∵$PQ$的中点在直线$l$上，
∴$PQ$中点的横坐标为$\frac{x_{1} + x_{2}}{2} = (-1) + 2 = 1$.
∴线段$PQ$的中点坐标为$(1,-1)$.]

考点三 直线与圆锥曲线的综合
例 5 (2025·昆明质检)已知双曲线$C:\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a＞0,b＞0)$的两个焦点分别为$F_{1}(-2,0)$，$F_{2}(2,0)$，点$P(5,\sqrt{23})$在双曲线$C$上.
(1)求双曲线$C$的方程；解 (1)依题意，$c = 2$，所以$a^{2} + b^{2} = 4$，则双曲线$C$的方程为$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{4 - a^{2}} = 1(0 ＜ a^{2} ＜ 4)$，将点$P(5,\sqrt{23})$代入上式，得$\frac{25}{a^{2}} - \frac{23}{4 - a^{2}} = 1$，解得$a^{2} = 50$(舍去)或$a^{2} = 2$，故所求双曲线的方程为$\frac{x^{2}}{2} - \frac{y^{2}}{2} = 1$.(2)依题意，可设直线$l$的方程为$y = kx + 2$，代入双曲线$C$的方程并整理，得$(1 - k^{2})x^{2} - 4kx - 6 = 0$.
(2)记$O$为坐标原点，过点$Q(0,2)$的直线$l$与双曲线$C$交于不同的两点$A$，$B$，若$\triangle OAB$的面积为$2\sqrt{2}$，求直线$l$的方程.因为直线$l$与双曲线$C$交于不同的两点$A$，$B$，所以$\begin{cases}1 - k^{2} \neq 0\\\Delta = (-4k)^{2} + 24(1 - k^{2}) ＞ 0\end{cases}$，解得$-\sqrt{3} ＜ k ＜ \sqrt{3}$且$k \neq \pm 1$.设$A(x_{1},y_{1})$，$B(x_{2},y_{2})$，则$x_{1} + x_{2} = \frac{4k}{1 - k^{2}}$，$x_{1}x_{2} = -\frac{6}{1 - k^{2}}$，所以$\left|AB\right| = \sqrt{1 + k^{2}} \cdot \sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}}=\sqrt{1 + k^{2}} \cdot \sqrt{(\frac{4k}{1 - k^{2}})^{2} + \frac{24}{1 - k^{2}}}$又原点$O$到直线$l$的距离$d = \frac{2}{\sqrt{1 + k^{2}}}$，所以$S_{\triangle OAB} = \frac{1}{2}d \cdot \left|AB\right| = \frac{1}{2} × \frac{2}{\sqrt{1 + k^{2}}} × \sqrt{1 + k^{2}} \cdot \sqrt{(\frac{4k}{1 - k^{2}})^{2} + \frac{24}{1 - k^{2}}}$又$S_{\triangle OAB} = 2\sqrt{2}$，即$\frac{\sqrt{3 - k^{2}}}{\left|1 - k^{2}\right|} = 1$，所以$k^{4} - k^{2} - 2 = 0$，解得$k = \pm \sqrt{2}$，满足条件.故满足条件的直线$l$有两条，其方程分别为$y = \sqrt{2}x + 2$和$y = -\sqrt{2}x + 2$.

答案： 例5 解
(1)依题意，$c = 2$，所以$a^{2} + b^{2} = 4$，则双曲线$C$的方程为$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{4 - a^{2}} = 1(0 ＜ a^{2} ＜ 4)$，将点$P(5,\sqrt{23})$代入上式，得$\frac{25}{a^{2}} - \frac{23}{4 - a^{2}} = 1$，解得$a^{2} = 50$(舍去)或$a^{2} = 2$，故所求双曲线的方程为$\frac{x^{2}}{2} - \frac{y^{2}}{2} = 1$.
(2)依题意，可设直线$l$的方程为$y = kx + 2$，代入双曲线$C$的方程并整理，得$(1 - k^{2})x^{2} - 4kx - 6 = 0$. 因为直线$l$与双曲线$C$交于不同的两点$A$，$B$，所以$\begin{cases}1 - k^{2} \neq 0\\\Delta = (-4k)^{2} + 24(1 - k^{2}) ＞ 0\end{cases}$，解得$-\sqrt{3} ＜ k ＜ \sqrt{3}$且$k \neq \pm 1$. 设$A(x_{1},y_{1})$，$B(x_{2},y_{2})$，则$x_{1} + x_{2} = \frac{4k}{1 - k^{2}}$，$x_{1}x_{2} = -\frac{6}{1 - k^{2}}$， 所以$\left|AB\right| = \sqrt{1 + k^{2}} \cdot \sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}}=\sqrt{1 + k^{2}} \cdot \sqrt{(\frac{4k}{1 - k^{2}})^{2} + \frac{24}{1 - k^{2}}}$ 又原点$O$到直线$l$的距离$d = \frac{2}{\sqrt{1 + k^{2}}}$， 所以$S_{\triangle OAB} = \frac{1}{2}d \cdot \left|AB\right| = \frac{1}{2} × \frac{2}{\sqrt{1 + k^{2}}} × \sqrt{1 + k^{2}} \cdot \sqrt{(\frac{4k}{1 - k^{2}})^{2} + \frac{24}{1 - k^{2}}}$ 又$S_{\triangle OAB} = 2\sqrt{2}$，即$\frac{\sqrt{3 - k^{2}}}{\left|1 - k^{2}\right|} = 1$，所以$k^{4} - k^{2} - 2 = 0$，解得$k = \pm \sqrt{2}$，满足条件. 故满足条件的直线$l$有两条，其方程分别为$y = \sqrt{2}x + 2$和$y = -\sqrt{2}x + 2$.

已知抛物线$C:y^{2}=3x$的焦点为$F$，斜率为$\frac{3}{2}$的直线$l$与$C$的交点为$A$，$B$，与$x$轴的交点为$P$.
(1)若$\vert AF\vert+\vert BF\vert=4$，求直线$l$的方程；
(2)若$\overrightarrow{AP}=3\overrightarrow{PB}$，求$\vert AB\vert$.解 设直线$l$的方程为$y = \frac{3}{2}x + t$，$A(x_{1},y_{1})$，$B(x_{2},y_{2})$.(1)由题设得$F(\frac{3}{4},0)$，故$\left|AF\right| + \left|BF\right| = x_{1} + x_{2} + \frac{3}{2}$.又$\left|AF\right| + \left|BF\right| = 4$，所以$x_{1} + x_{2} = \frac{5}{2}$.由$\begin{cases}y = \frac{3}{2}x + t\\y^{2} = 3x\end{cases}$可得$9x^{2} + 12(t - 1)x + 4t^{2} = 0$，其中$\Delta = 144(1 - 2t) ＞ 0$，则$x_{1} + x_{2} = -\frac{12(t - 1)}{9}$，从而$-\frac{12(t - 1)}{9} = \frac{5}{2}$，得$t = -\frac{7}{8}$(满足$\Delta ＞ 0$)，所以$l$的方程为$y = \frac{3}{2}x - \frac{7}{8}$.(2)由$\overrightarrow{AP} = 3\overrightarrow{PB}$，可得$y_{1} = -3y_{2}$.由$\begin{cases}y = \frac{3}{2}x + t\\y^{2} = 3x\end{cases}$可得$y^{2} - 2y + 2t = 0$，其中$\Delta = 4 - 8t ＞ 0$，所以$y_{1} + y_{2} = 2$，从而$-3y_{2} + y_{2} = 2$，故$y_{2} = -1$，$y_{1} = 3$.代入$C$的方程得$x_{1} = 3$，$x_{2} = \frac{1}{3}$.所以$A(3,3)$，$B(\frac{1}{3},-1)$，故$\left|AB\right| = \frac{4\sqrt{13}}{3}$.

答案： 训练3 解 设直线$l$的方程为$y = \frac{3}{2}x + t$，$A(x_{1},y_{1})$，$B(x_{2},y_{2})$.
(1)由题设得$F(\frac{3}{4},0)$，故$\left|AF\right| + \left|BF\right| = x_{1} + x_{2} + \frac{3}{2}$. 又$\left|AF\right| + \left|BF\right| = 4$，所以$x_{1} + x_{2} = \frac{5}{2}$.由$\begin{cases}y = \frac{3}{2}x + t\\y^{2} = 3x\end{cases}$可得$9x^{2} + 12(t - 1)x + 4t^{2} = 0$，其中$\Delta = 144(1 - 2t) ＞ 0$， 则$x_{1} + x_{2} = -\frac{12(t - 1)}{9}$，从而$-\frac{12(t - 1)}{9} = \frac{5}{2}$，得$t = -\frac{7}{8}$(满足$\Delta ＞ 0$)，所以$l$的方程为$y = \frac{3}{2}x - \frac{7}{8}$.
(2)由$\overrightarrow{AP} = 3\overrightarrow{PB}$，可得$y_{1} = -3y_{2}$. 由$\begin{cases}y = \frac{3}{2}x + t\\y^{2} = 3x\end{cases}$可得$y^{2} - 2y + 2t = 0$，其中$\Delta = 4 - 8t ＞ 0$， 所以$y_{1} + y_{2} = 2$，从而$-3y_{2} + y_{2} = 2$，故$y_{2} = -1$，$y_{1} = 3$. 代入$C$的方程得$x_{1} = 3$，$x_{2} = \frac{1}{3}$. 所以$A(3,3)$，$B(\frac{1}{3},-1)$，故$\left|AB\right| = \frac{4\sqrt{13}}{3}$.