《2025年高考帮数学》

例1　[2023全国卷乙]已知椭圆$C:\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}} = 1(a＞b＞0)$的离心率为$\frac{\sqrt{5}}{3}$，点$A(-2,0)$在$C$上.
(1)求$C$的方程；
(2)过点$(-2,3)$的直线交$C$于$P,Q$两点，直线$AP,AQ$与$y$轴的交点分别为$M,N$，证明：线段$MN$的中点为定点.

答案：
(1)因为点A(-2,0)在C上,所以b = 2.
因为椭圆的离心率e=$\frac{c}{a}=\sqrt{1 - \frac{b^{2}}{a^{2}}}=\frac{\sqrt{5}}{3}$,所以$a^{2}=9$,故椭圆C的方程为$\frac{y^{2}}{9}+\frac{x^{2}}{4}=1$.
(2)由题意知,直线PQ的斜率存在且不为0,设$l_{PQ}:y - 3 = k(x + 2)$,$P(x_{1},y_{1})$,$Q(x_{2},y_{2})$,
由$\begin{cases}y - 3 = k(x + 2)\\\frac{y^{2}}{9}+\frac{x^{2}}{4}=1\end{cases}$,得$(4k^{2}+9)x^{2}+(16k^{2}+24k)x + 16k^{2}+48k = 0$,
则$\Delta=(16k^{2}+24k)^{2}-4(4k^{2}+9)(16k^{2}+48k)=-36\times48k＞0$,
故$x_{1}+x_{2}=-\frac{16k^{2}+24k}{4k^{2}+9}$,$x_{1}x_{2}=\frac{16k^{2}+48k}{4k^{2}+9}$.
直线AP:$y=\frac{y_{1}}{x_{1}+2}(x + 2)$,令x = 0,解得$y_{M}=\frac{2y_{1}}{x_{1}+2}$,同理得$y_{N}=\frac{2y_{2}}{x_{2}+2}$,
则$y_{M}+y_{N}=2\times\frac{y_{1}(x_{2}+2)+y_{2}(x_{1}+2)}{(x_{1}+2)(x_{2}+2)}$
$=2\times\frac{(kx_{1}+2k + 3)(x_{2}+2)+(kx_{2}+2k + 3)(x_{1}+2)}{(x_{1}+2)(x_{2}+2)}$
$=2\times\frac{2kx_{1}x_{2}+(4k + 3)(x_{1}+x_{2})+8k + 12}{x_{1}x_{2}+2(x_{1}+x_{2})+4}$
$=2\times\frac{2k(16k^{2}+48k)+(4k + 3)(-16k^{2}-24k)+(8k + 12)(4k^{2}+9)}{16k^{2}+48k+2(-16k^{2}-24k)+4(4k^{2}+9)}$
$=2\times\frac{108}{36}$
= 6.
所以MN的中点的纵坐标为$\frac{y_{M}+y_{N}}{2}=3$,所以MN的中点为定点(0,3).

训练1　[2022全国卷乙]已知椭圆$E$的中心为坐标原点，对称轴为$x$轴、$y$轴，且过$A(0,-2)$，$B(\frac{3}{2},-1)$两点.
(1)求$E$的方程；
(2)设过点$P(1,-2)$的直线交$E$于$M,N$两点，过$M$且平行于$x$轴的直线与线段$AB$交于点$T$，点$H$满足$\overrightarrow{MT}=\overrightarrow{TH}$. 证明：直线$HN$过定点.

答案：
(1)
∵椭圆E的中心为坐标原点，对称轴为x轴、y轴，且过A(0,-2),
∴可设椭圆E的方程为$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1$,又椭圆E过$B(\frac{3}{2},-1)$,
∴$\frac{9}{4a^{2}}+\frac{1}{4}=1$,得$a^{2}=3$,
∴E的方程为$\frac{x^{2}}{3}+\frac{y^{2}}{4}=1$.
(2)当直线MN的斜率不存在时,$l_{MN}:x = 1$,
由$\begin{cases}x = 1\\\frac{x^{2}}{3}+\frac{y^{2}}{4}=1\end{cases}$,得$y^{2}=\frac{8}{3}$,
∴$y=\pm\frac{2\sqrt{2}}{\sqrt{3}}$.
结合题意可知$M(1,-\frac{2\sqrt{2}}{\sqrt{3}})$,$N(1,\frac{2\sqrt{2}}{\sqrt{3}})$,
∴过M且平行于x轴的直线的方程为$y = -\frac{2\sqrt{2}}{\sqrt{3}}$.
易知点T的横坐标$x_{T}\in[0,\frac{3}{2}]$,直线AB的方程为$y - (-2)=\frac{-1 - (-2)}{\frac{3}{2}-0}\times(x - 0)$,即$y=\frac{2}{3}x - 2$,
由$\begin{cases}y = -\frac{2\sqrt{2}}{\sqrt{3}}\\y=\frac{2}{3}x - 2\end{cases}$,得$x_{T}=3-\sqrt{6}$,
∴$T(3-\sqrt{6},-\frac{2\sqrt{2}}{\sqrt{3}})$.
∵$\overrightarrow{MT}=\overrightarrow{TH}$,
∴$H(5 - 2\sqrt{6},-\frac{2\sqrt{2}}{\sqrt{3}})$,
$l_{HN}:y-\frac{2\sqrt{2}}{\sqrt{3}}=\frac{\frac{4\sqrt{2}}{\sqrt{3}}}{2\sqrt{6}-4}(x - 1)$,即$y=\frac{2(3+\sqrt{6})}{3}x - 2$.
易知直线HN过定点(0,-2).
当直线MN的斜率存在时，如图，设$M(x_{1},y_{1})$,$N(x_{2},y_{2})$,$l_{MN}:y = kx + m$ ($k + m=-2$).
由$\begin{cases}y = kx + m\\\frac{x^{2}}{3}+\frac{y^{2}}{4}=1\end{cases}$,得$(3k^{2}+4)x^{2}+6kmx + 3m^{2}-12 = 0$,$\Delta＞0$,
∴$x_{1}+x_{2}=-\frac{6km}{3k^{2}+4}$,$x_{1}x_{2}=\frac{3m^{2}-12}{3k^{2}+4}$.
过M且平行于x轴的直线的方程为$y = y_{1}$,
与直线AB的方程联立，得$\begin{cases}y = y_{1}\\y=\frac{2}{3}x - 2\end{cases}$,得$x_{T}=\frac{3(y_{1}+2)}{2}$,
∴$T(\frac{3(y_{1}+2)}{2},y_{1})$.
∵$\overrightarrow{MT}=\overrightarrow{TH}$,
∴$H(3y_{1}+6 - x_{1},y_{1})$,
$l_{HN}:y - y_{2}=\frac{y_{1}-y_{2}}{3y_{1}+6 - x_{1}-x_{2}}(x - x_{2})$,
即$y=\frac{y_{1}-y_{2}}{3y_{1}+6 - x_{1}-x_{2}}x + y_{2}-\frac{y_{1}-y_{2}}{3y_{1}+6 - x_{1}-x_{2}}\cdot x_{2}$.
令x = 0,得$y = y_{2}-\frac{(y_{1}-y_{2})x_{2}}{3y_{1}+6 - x_{1}-x_{2}}=\frac{-(x_{1}y_{2}+x_{2}y_{1})+3y_{1}y_{2}+6y_{2}}{-(x_{1}+x_{2})+6 + 3y_{1}}$
$=\frac{-(x_{1}y_{2}+x_{2}y_{1})+3y_{1}y_{2}+6y_{2}}{-(x_{1}+x_{2})+6 + 3(y_{1}+y_{2})-3y_{2}}$.
∵$y_{1}y_{2}=(kx_{1}+m)(kx_{2}+m)=k^{2}x_{1}x_{2}+mk(x_{1}+x_{2})+m^{2}=\frac{-12k^{2}+4m^{2}}{3k^{2}+4}$,$y_{1}+y_{2}=(kx_{1}+m)+(kx_{2}+m)=k(x_{1}+x_{2})+2m=\frac{8m}{3k^{2}+4}$,$x_{1}y_{2}+x_{2}y_{1}=x_{1}(kx_{2}+m)+x_{2}(kx_{1}+m)=2kx_{1}x_{2}+m(x_{1}+x_{2})=\frac{-24k}{3k^{2}+4}$,
∴$-(x_{1}y_{2}+x_{2}y_{1})+3y_{1}y_{2}=\frac{24k}{3k^{2}+4}+\frac{-36k^{2}+12m^{2}}{3k^{2}+4}=\frac{-36k^{2}+12m^{2}+24k}{3k^{2}+4}=\frac{-24(k^{2}-3k - 2)}{3k^{2}+4}$,
$-(x_{1}+x_{2})+6 + 3(y_{1}+y_{2})=\frac{6km}{3k^{2}+4}+6+\frac{24m}{3k^{2}+4}=\frac{6km + 18k^{2}+24 + 24m}{3k^{2}+4}=\frac{12(k^{2}-3k - 2)}{3k^{2}+4}$,
∴$y=\frac{\frac{-24(k^{2}-3k - 2)}{3k^{2}+4}+6y_{2}}{\frac{12(k^{2}-3k - 2)}{3k^{2}+4}-3y_{2}}=-2$,
∴直线HN过定点(0,-2).
综上,直线HN过定点(0,-2).