《2025年世纪金榜高中全程复习方略高中数学A版》

[例2]已知椭圆$C:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a＞b＞0)$的左、右焦点分别为$F_1(-c,0)$和$F_2(c,0)$，离心率是$\frac{\sqrt{3}}{2}$，直线$x = c$被椭圆截得的弦长等于2.
(1)求椭圆$C$的标准方程；
(2)若直线$l:x + 2y - 2 = 0$与椭圆相交于$A,B$两点，$O$为坐标原点，求$\triangle OAB$的面积.

答案： 【解析】
(1)由题意可得$\begin{cases}\frac{c}{a}=\frac{\sqrt{3}}{2}\\\frac{2b^{2}}{a}=2\\a^{2}=b^{2}+c^{2}\end{cases}$，解得$\begin{cases}a = 4\\b = 2\end{cases}$，所以椭圆$C$的标准方程为$\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$；
(2)设直线$l$与椭圆的交点为$A(x_{1},y_{1})$，$B(x_{2},y_{2})$，联立$\begin{cases}x + 2y - 2 = 0\\\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\end{cases}$，消元整理可得$x^{2}-2x - 6 = 0$，显然$\Delta＞0$，故$x_{1}+x_{2}=2$，$x_{1}x_{2}=-6$，所以$|AB|=\sqrt{1 + k^{2}}|x_{1}-x_{2}|=\sqrt{1 + k^{2}}\cdot\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\sqrt{1+\frac{1}{4}}\cdot\sqrt{2^{2}+4\times6}=\sqrt{35}$，又由于原点到直线$l$的距离为$d=\frac{2}{\sqrt{1 + 2^{2}}}=\frac{2}{\sqrt{5}}$，所以$S_{\triangle OAB}=\frac{1}{2}\cdot d\cdot|AB|=\frac{1}{2}\times\frac{2}{\sqrt{5}}\times\sqrt{35}=\sqrt{7}$，所以$\triangle OAB$的面积为$\sqrt{7}$.

[例3](一题多法)(2022·新高考Ⅱ卷)已知直线$l$与椭圆$\frac{x^{2}}{6}+\frac{y^{2}}{3}=1$在第一象限交于$A,B$两点，$l$与$x$轴、$y$轴分别交于$M,N$两点，且$|MA| = |NB|,|MN| = 2\sqrt{3}$，则$l$的方程为________.

答案： 【解析】方法一：设直线$l$的方程为$\frac{x}{m}+\frac{y}{n}=1(m＞0,n＞0)$，分别令$y = 0$，$x = 0$，得点$M(m,0)$，$N(0,n)$.设$A(x_{1},y_{1})$，$B(x_{2},y_{2})$.由题意知线段$AB$与线段$MN$有相同的中点，所以$\begin{cases}\frac{x_{1}+x_{2}}{2}=\frac{m + 0}{2}\\\frac{y_{1}+y_{2}}{2}=\frac{0 + n}{2}\end{cases}$，即$\begin{cases}x_{1}+x_{2}=m\\y_{1}+y_{2}=n\end{cases}$.因为$k_{AB}=k_{MN}$，所以$\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{0 - n}{m - 0}=-\frac{n}{m}$.将$A(x_{1},y_{1})$，$B(x_{2},y_{2})$代入椭圆方程，得$\begin{cases}\frac{x_{1}^{2}}{6}+\frac{y_{1}^{2}}{3}=1\\\frac{x_{2}^{2}}{6}+\frac{y_{2}^{2}}{3}=1\end{cases}$，相减得$\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{6}+\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{3}=0$，由题意知$x_{1}+x_{2}\neq0$，$x_{1}\neq x_{2}$，所以$\frac{y_{1}+y_{2}}{x_{1}+x_{2}}\cdot\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\frac{1}{2}$，即$\frac{n}{m}\cdot(-\frac{n}{m})=-\frac{1}{2}$，整理得$m^{2}=2n^{2}$.①又$|MN| = 2\sqrt{3}$，所以由勾股定理，得$m^{2}+n^{2}=12$.②由①②并结合$m＞0$，$n＞0$，得$\begin{cases}m = 2\sqrt{2}\\n = 2\end{cases}$，所以直线$l$的方程为$\frac{x}{2\sqrt{2}}+\frac{y}{2}=1$，即$x+\sqrt{2}y - 2\sqrt{2}=0$.
方法二：设直线$l$的方程为$\frac{x}{m}+\frac{y}{n}=1(m＞0,n＞0)$，分别令$y = 0$，$x = 0$，得点$M(m,0)$，$N(0,n)$.由题意知线段$AB$与线段$MN$有相同的中点，设为$Q$，则$Q(\frac{m}{2},\frac{n}{2})$，则$k_{AB}=\frac{0 - n}{m - 0}=-\frac{n}{m}$，$k_{OQ}=\frac{\frac{n}{2}}{\frac{m}{2}}=\frac{n}{m}$.由椭圆中点弦的性质知，$k_{AB}\cdot k_{OQ}=-\frac{b^{2}}{a^{2}}=-\frac{1}{2}$，即$(-\frac{n}{m})\cdot\frac{n}{m}=-\frac{1}{2}$，以下同方法一.
答案：$x+\sqrt{2}y - 2\sqrt{2}=0$

1.(2024·南昌模拟)已知直线$l$交椭圆$C:\frac{y^{2}}{9}+\frac{x^{2}}{4}=1$于$A,B$两点，若点$M(1,2)$为$A,B$两点的中点，则直线$l$的斜率为（ ）
A.$\frac{2}{9}$
B.$-\frac{2}{9}$
C.$\frac{9}{8}$
D.$-\frac{9}{8}$

答案： D 椭圆$C:\frac{y^{2}}{9}+\frac{x^{2}}{4}=1$，依题意可知直线$l$的斜率存在，设$A(x_{1},y_{1})$，$B(x_{2},y_{2})$，则$\begin{cases}\frac{y_{1}^{2}}{9}+\frac{x_{1}^{2}}{4}=1\\\frac{y_{2}^{2}}{9}+\frac{x_{2}^{2}}{4}=1\end{cases}$，两式相减并化简得$-\frac{9}{4}=\frac{y_{1}+y_{2}}{x_{1}+x_{2}}\cdot\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$，即$-\frac{9}{4}=\frac{4}{2}\cdot\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$，$\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\frac{9}{8}$，所以直线$l$的斜率为$-\frac{9}{8}$.

2. 若椭圆$\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$的弦$AB$的中点为$(-1,-1)$，则弦$AB$的长为（ ）
A.$\frac{\sqrt{30}}{3}$
B.$\frac{2\sqrt{6}}{3}$
C.$\frac{\sqrt{10}}{3}$
D.$\frac{\sqrt{15}}{3}$

答案： A 设$A(x_{1},y_{1})$，$B(x_{2},y_{2})$，因为弦$AB$的中点为$(-1,-1)$，可得$x_{1}+x_{2}=-2$，$y_{1}+y_{2}=-2$，又因为$A$，$B$在椭圆上，可得$\begin{cases}\frac{x_{1}^{2}}{4}+\frac{y_{1}^{2}}{2}=1\\\frac{x_{2}^{2}}{4}+\frac{y_{2}^{2}}{2}=1\end{cases}$，两式相减可得$\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{4}+\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{2}=0$，可得$\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\frac{2\cdot(x_{1}+x_{2})}{4\cdot(y_{1}+y_{2})}=-\frac{1}{2}$，即直线$AB$的斜率为$k = -\frac{1}{2}$，所以弦$AB$的直线方程为$y + 1=-\frac{1}{2}(x + 1)$，即$y = -\frac{1}{2}x-\frac{3}{2}$，联立$\begin{cases}y = -\frac{1}{2}x-\frac{3}{2}\\\frac{x^{2}}{4}+\frac{y^{2}}{2}=1\end{cases}$，整理得$3x^{2}+6x + 1 = 0$，可得$x_{1}+x_{2}=-2$，$x_{1}x_{2}=\frac{1}{3}$，由弦长公式，可得$|AB|=\sqrt{1+(-\frac{1}{2})^{2}}\cdot\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\sqrt{\frac{5}{4}}\cdot\sqrt{(-2)^{2}-4\times\frac{1}{3}}=\frac{\sqrt{30}}{3}$.