简单直白的面积问题

已知椭圆\(C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左定点合右焦点分别为\(Q,F\)，且\(|QF|=3\)，点\(D(0,1)\)满足\(\overrightarrow{DQ}\cdot\overrightarrow{DF}=-1\)

(1)求\(C\)方程

(2)过点\(D\)的直线\(l\)与\(C\)交于点\(A,B\)两点，与\(x\)轴交于点\(T\)，且点\(T\)在\(Q\)点左侧，点\(B\)关于\(x\)轴的对称点为\(E\)，直线\(QA,QE\)分别与直线\(1\)交于点\(M,N\)两点，求\(\triangle TMN\)面积的最小值

解

(1)\(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\)

(2)\(S_{\triangle TMN}=\dfrac{1}{2}\left(1-x_T\right)(y_{M}-y_{N})\)

设\(AB:y=kx+m,A(x_1,y_1),B(x_2,y_2),E(x_2,-y_2),x_T=-\dfrac{m}{k}\)

则有\(AQ:y=\dfrac{y_1}{x_1+2}(x+2),y_M=\dfrac{3y_1}{x_1+2}\)

\(QE:y=\dfrac{-y_2}{x_2+2}(x+2),y_N=\dfrac{-3y_2}{x_2+2}\)

\(y_M-y_N=\dfrac{3y_1}{x_1+2}+\dfrac{3y_2}{x_2+2}\)

\(=3\cdot\dfrac{(kx_1+m)(x_2+2)+(kx_2+m)(x_1+2)}{x_1x_2+2(x_1+x_2)+4}\)

\(=3\cdot\dfrac{2kx_1x_2+2k(x_1+x_2)+m(x_1+x_2)+4m}{x_1x_2+2(x_1+x_2)+4}\)

联立\(\begin{cases} y=kx+m\\ \dfrac{x^2}{4}+\dfrac{y^2}{3}=1 \end{cases}\)有\((3+4k^2)x^2+8kmx+4m^2-12=0\)

有\(\begin{cases} x_1+x_2=-\dfrac{8km}{3+4k^2}\\ x_1x_2=\dfrac{4m^2-12}{3+4k^2} \end{cases}\)

则
\(S_{\triangle TMN}=\dfrac{1}{2}\left(1-x_T\right)(y_{M}-y_{N})\)

\(=\dfrac{1}{2}\left(1+\dfrac{m}{k}\right)\cdot 3\cdot \dfrac{2kx_1x_2+2k(x_1+x_2)+m(x_1+x_2)+4m}{x_1x_2+2(x_1+x_2)+4}\)

\(=\dfrac{3}{2}\left(1+\dfrac{m}{k}\right)\cdot\dfrac{8km^2-24k-16k^2m-8km^2+12m+16k^2m}{4m^2-12-16km+12+16k^2}\)

\(=\dfrac{3}{2}\left(1+\dfrac{m}{k}\right)\cdot\dfrac{12m-24k}{4m^2-16km+16k^2}\)

\(=\dfrac{3}{2}\left(1+\dfrac{m}{k}\right)\dfrac{3(m-2k)}{(m-2k)^2}\)

\(=\dfrac{9}{2}\left(1+\dfrac{m}{k}\right)\dfrac{1}{m-2k}\)

\(=\dfrac{9}{2}\cdot\dfrac{k+m}{k(m-2k)}\)

因\(m=1\)，则

\(S=\dfrac{9}{2}\cdot\dfrac{k+1}{k(1-2k)}\)

\(\xlongequal[]{k+1=t}\dfrac{9}{2}\cdot\dfrac{t}{(t-1)(3-2t)}\)

\(=\dfrac{9}{2}\cdot\dfrac{t}{-2t^2+5t-3}\)

\(=\dfrac{9}{2}\cdot\dfrac{1}{5-\left(\dfrac{3}{t}+2t\right)}\)

\(\geq \dfrac{9}{2}\left(5+2\sqrt{6}\right)\)