《2025年成才之路高中新课程学习指导高中数学选择性必修第一册北师大版》

抛物线 $y^2 = 12x$ 截直线 $y = 2x + 1$ 所得弦长等于 (A)

A.$\sqrt{15}$
B.$2\sqrt{15}$
C.$\frac{\sqrt{15}}{2}$
D.$15$

答案： 令直线与抛物线交于点$A(x_{1},y_{1}),B(x_{2},y_{2})$，
由$\begin{cases}y = 2x + 1,\\y^{2} = 12x,\end{cases}$得$4x^{2} - 8x + 1 = 0$，$\therefore x_{1} + x_{2} = 2$，$x_{1}x_{2} = \frac{1}{4}$，
$\therefore\vert AB\vert = \sqrt{(1 + 2^{2})(x_{1} - x_{2})^{2}}=\sqrt{5[(x_{1} + x_{2})^{2} - 4x_{1}x_{2}]} = \sqrt{15}$.

例 2. 已知椭圆 $\frac{x^2}{16} + \frac{y^2}{4} = 1$，求：
(1) 以 $P(2, -1)$ 为中点的弦所在直线的方程；
(2) 斜率为 $2$ 的平行弦中点的轨迹方程；
(3) 过 $Q(8, 2)$ 的直线被椭圆截得的弦的中点的轨迹方程。

答案： 设弦的两端点分别为$A(x_{1},y_{1}),B(x_{2},y_{2})$，$AB$中点为$R(x,y)$，则$2x = x_{1} + x_{2}$，$2y = y_{1} + y_{2}$.
又$A$，$B$两点均在椭圆上，
故有$x_{1}^{2} + 4y_{1}^{2} = 16$，$x_{2}^{2} + 4y_{2}^{2} = 16$.
两式相减，得$(x_{1} + x_{2})(x_{1} - x_{2}) = - 4(y_{1} + y_{2})(y_{1} - y_{2})$.
故$k_{AB} = \frac{y_{1} - y_{2}}{x_{1} - x_{2}} = - \frac{x_{1} + x_{2}}{4(y_{1} + y_{2})} = - \frac{x}{4y}$
(1)由$k_{AB} = - \frac{x}{4y} = \frac{1}{2}$，得所求轨迹方程为$x - 2y - 4 = 0$.
(2)由$k_{AB} = - \frac{x}{4y} = 2$，得所求轨迹方程为$x + 8y = 0(-4 \leq x \leq 4)$.
(3)由$k_{AB} = - \frac{x}{4y} = \frac{y - 2}{x - 8}$，得所求轨迹方程为$(x - 4)^{2} + 4(y - 1)^{2} = 20( - 4 \leq x \leq 4)$.

已知椭圆 $x^2 + 2y^2 = 4$，则以 $(1, 1)$ 为中点的弦的长度为 (C)

A.$3\sqrt{2}$
B.$2\sqrt{3}$
C.$\frac{\sqrt{30}}{3}$
D.$\frac{3\sqrt{6}}{2}$

答案： 依题设弦端点$A(x_{1},y_{1})$，$B(x_{2},y_{2})$，
则$x_{1} + x_{2} = 2$，$y_{1} + y_{2} = 2$，
又$x_{1}^{2} + 2y_{1}^{2} = 4$，$x_{2}^{2} + 2y_{2}^{2} = 4$，
$\therefore x_{1}^{2} - x_{2}^{2} = - 2(y_{1}^{2} - y_{2}^{2})$，
此弦的斜率$k = \frac{y_{1} - y_{2}}{x_{1} - x_{2}} = \frac{x_{1} + x_{2}}{-2(y_{1} + y_{2})} = - \frac{1}{2}$，
$\therefore$此弦所在的直线方程为$y - 1 = - \frac{1}{2}(x - 1)$，
即$y = - \frac{1}{2}x + \frac{3}{2}$.
代入$x^{2} + 2y^{2} = 4$，整理得$3x^{2} - 6x + 1 = 0$，
$\therefore x_{1}x_{2} = \frac{1}{3}$，$x_{1} + x_{2} = 2$，
$\therefore\vert AB\vert = \sqrt{1 + k^{2}} · \sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}}=\sqrt{1 + \frac{1}{4}} × \sqrt{4 - 4 × \frac{1}{3}} = \frac{\sqrt{30}}{3}$.

例 3. 在平面直角坐标系 $xOy$ 中，已知椭圆 $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a ＞ b ＞ 0)$ 的焦距为 $2$，离心率为 $\frac{\sqrt{2}}{2}$，椭圆的右顶点为 $A$。
(1) 求椭圆的标准方程；
(2) 过点 $D(\sqrt{2}, -\sqrt{2})$ 作直线交椭圆于两个不同点 $P, Q$，求证：直线 $AP, AQ$ 的斜率之和为定值。
[分析] （1）由条件求出 $a$ 与 $b$ 的值，从而得出椭圆的标准方程；（2）$A$ 点坐标已知，设出点 $P, Q$ 的坐标并建立其关系，表示出斜率并求和，通过恒等变形即得证。

答案：
(1)因为焦距$2c = 2$，所以$c = 1$，又椭圆的离心率$e = \frac{c}{a} = \frac{\sqrt{2}}{2}$，则$a = \sqrt{2}$，所以$b^{2} = a^{2} - c^{2} = 1$，
故椭圆的标准方程为$\frac{x^{2}}{2} + y^{2} = 1$.
(2)证明:易知$A(\sqrt{2},0)$，设$P(x_{1},y_{1})$，$Q(x_{2},y_{2})$，点$D$在点$A$正下方，过点$D$作垂直于$x$轴的直线，所作直线与椭圆只有一个交点，不合题意，于是直线$PQ$的斜率存在，设直线$PQ$的方程为$y = k(x - \sqrt{2}) - \sqrt{2}$，
由$\begin{cases}\frac{x^{2}}{2} + y^{2} = 1,\\y = k(x - \sqrt{2}) - \sqrt{2}\end{cases}$消去$y$并化简得$(2k^{2} + 1)x^{2} - (4\sqrt{2}k^{2} + 4\sqrt{2}k)x + 4k^{2} + 8k + 2 = 0$，则$x_{1} + x_{2} = \frac{4\sqrt{2}k^{2} + 4\sqrt{2}k}{2k^{2} + 1}$，$x_{1}x_{2} = \frac{4k^{2} + 8k + 2}{2k^{2} + 1}$，
所以$y_{1} + y_{2} = k(x_{1} + x_{2}) - 2\sqrt{2}k - 2\sqrt{2} = \frac{-2\sqrt{2} - 2\sqrt{2}k}{2k^{2} + 1}$
所以$k_{AP} + k_{AQ} = \frac{y_{1}}{x_{1} - \sqrt{2}} + \frac{y_{2}}{x_{2} - \sqrt{2}}=\frac{y_{1}x_{2} + y_{2}x_{1} - \sqrt{2}(y_{1} + y_{2})}{x_{1}x_{2} - \sqrt{2}(x_{1} + x_{2}) + 2}$，
又$y_{1}x_{2} + y_{2}x_{1} = [k(x_{1} - \sqrt{2}) - \sqrt{2}]x_{2} + [k(x_{2} - \sqrt{2}) - \sqrt{2}]x_{1}= 2kx_{1}x_{2} - (\sqrt{2}k + \sqrt{2})(x_{1} + x_{2}) = \frac{-4k}{2k^{2} + 1}$，
所以$k_{AP} + k_{AQ} = \frac{\frac{-4k}{2k^{2} + 1} - \sqrt{2} × \frac{-2\sqrt{2} - 2\sqrt{2}k}{2k^{2} + 1}}{\frac{4k^{2} + 8k + 2}{2k^{2} + 1} - \sqrt{2} × \frac{4\sqrt{2}k^{2} + 4\sqrt{2}k}{2k^{2} + 1} + 2} = 1$.故直线$AP$，$AQ$的斜率之和为定值$1$.