21．（12分）已知椭圆$E$的中心为坐标原点，对称轴为$x$轴、$y$轴，且过$A(0,-2)$，$B(\dfrac{3}{2}$，$-1)$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．
分析：（1）设$E$的方程为$mx^{2}+ny^{2}=1(m > 0$，$n > 0$且$m\ne n)$，将$A$，$B$两点坐标代入即可求解；（2）由$A(0,-2),B(\dfrac{3}{2},-1)$可得线段$AB:y=\dfrac{2}{3}x-2$，①若过$P(1,-2)$的直线的斜率不存在，直线为$x=1$，代入椭圆方程，根据$\overrightarrow{MT}=\overrightarrow{TH}$即可求解；②若过$P(1,-2)$的直线的斜率存在，设$kx-y-(k+2)=0$，$M(x_{1}$，$y_{1})$，$N(x_{2}$，$y_{2})$，联立$\left\{\begin{array}{l}kx-y-(k+2)=0\\  \dfrac{x^{2}}{3}+\dfrac{y^{2}}{4}=1\end{array}\right.$，得$(3k^{2}+4)x^{2}-6k(2+k)x+3k(k+4)=0$，结合韦达定理和已知条件即可求解．
解：（1）设$E$的方程为$mx^{2}+ny^{2}=1(m > 0$，$n > 0$且$m\ne n)$，
将$A(0,-2),B(\dfrac{3}{2},-1)$两点代入得$\left\{\begin{array}{l}{4n=1}\\ {\dfrac{9}{4}m+n=1}\end{array}\right.$，
解得$m=\dfrac{1}{3}$，$n=\dfrac{1}{4}$，
故$E$的方程为$\dfrac{x^{2}}{3}+\dfrac{y^{2}}{4}=1$；
（2）由$A(0,-2),B(\dfrac{3}{2},-1)$可得线段$AB:y=\dfrac{2}{3}x-2$
（1）若过点$P(1,-2)$的直线斜率不存在，直线$x=1$．代入$\dfrac{x^{2}}{3}+\dfrac{y^{2}}{4}=1$，
可得$M(1,-\dfrac{2\sqrt{6}}{3})$，$N(1,\dfrac{2\sqrt{6}}{3})$，将$y=-\dfrac{2\sqrt{6}}{3}$代入$y=\dfrac{2}{3}x-2$，可得$T(-\sqrt{6}+3,-\dfrac{2\sqrt{6}}{3})$，得到$H(-2\sqrt{6}+5$，$-\dfrac{2\sqrt{6}}{3})$求得$HN$ 方程：$y=(2+\dfrac{2\sqrt{6}}{3})x-2$，过点$(0,-2)$．
②若过$P(1,-2)$的直线的斜率存在，设$kx-y-(k+2)=0$，$M(x_{1}$，$y_{1})$，$N(x_{2}$，$y_{2})$，
联立$\left\{\begin{array}{l}kx-y-(k+2)=0\\  \dfrac{x^{2}}{3}+\dfrac{y^{2}}{4}=1\end{array}\right.$，得$(3k^{2}+4)x^{2}-6k(2+k)x+3k(k+4)=0$，
故有$\left\{\begin{array}{l}x_{1}+x_{2}=\dfrac{6k(2+k)}{3k^{2}+4}\\  x_{1}x_{2}=\dfrac{3k(4+k)}{3k^{2}+4}\end{array}\right.$，$\left\{\begin{array}{l}y_{1}+y_{2}=\dfrac{-8(2+k)}{3k^{2}+4}\\  y_{1}y_{2}=\dfrac{4(4+4k-2k^{2})}{3k^{2}+4}\end{array}\right.$，
$x_{1}y_{2}+x_{2}y_{1}=x_{1}(kx_{2}-k-2)+x_{2}(kx_{1}-k-2)$
$=kx_{1}x_{2}-kx_{1}-2x_{1}+kx_{1}x_{2}-kx_{2}-2x_{2}$
$=2kx_{1}x_{2}-(k+2)(x_{1}+x_{2})$
$=2k\cdot \dfrac{3k(4+k)}{3{k}^{2}+4}-(k+2)\cdot \dfrac{6k(2+k)}{3{k}^{2}+4}$
$=\dfrac{-24k}{3{k}^{2}+4}$，
$\therefore$$x_{1}y_{2}+x_{2}y_{1}=\dfrac{-24k}{3k^{2}+4}(*)$，
联立$\left\{\begin{array}{l}y=y_{1}\\  y=\dfrac{2}{3}x-2\end{array}\right.$，可得$T(\dfrac{3y_{1}}{2}+3,y_{1}),H(3y_{1}+6-x_{1},y_{1})$，
可求得此时$HN:y-y_{2}=\dfrac{y_{1}-y_{2}}{3y_{1}+6-x_{1}-x_{2}}(x-x_{2})$，
将$(0,-2)$代入整理得$2(x_{1}+x_{2})-6(y_{1}+y_{2})+x_{1}y_{2}+x_{2}y_{1}-3y_{1}y_{2}-12=0$，
将$(*)$代入，得$24k+12k^{2}+96+48k-24k-48-48k+24k^{2}-36k^{2}-48=0$，
显然成立．
综上，可得直线$HN$过定点$(0,-2)$．
点评：本题考查了直线与椭圆的综合应用，属于中档题．