簡單直白的面積問題
已知橢圓\(C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左定點合右焦點分別為\(Q,F\)，且\(|QF|=3\)，點\(D(0,1)\)滿足\(\overrightarrow{DQ}\cdot\overrightarrow{DF}=-1\)
(1)求\(C\)方程
(2)過點\(D\)的直線\(l\)與\(C\)交於點\(A,B\)兩點，與\(x\)軸交於點\(T\)，且點\(T\)在\(Q\)點左側，點\(B\)關於\(x\)軸的對稱點為\(E\)，直線\(QA,QE\)分別與直線\(1\)交於點\(M,N\)兩點，求\(\triangle TMN\)面積的最小值
解
(1)\(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\)
(2)\(S_{\triangle TMN}=\dfrac{1}{2}\left(1-x_T\right)(y_{M}-y_{N})\)
設\(AB:y=kx+m,A(x_1,y_1),B(x_2,y_2),E(x_2,-y_2),x_T=-\dfrac{m}{k}\)
則有\(AQ:y=\dfrac{y_1}{x_1+2}(x+2),y_M=\dfrac{3y_1}{x_1+2}\)
\(QE:y=\dfrac{-y_2}{x_2+2}(x+2),y_N=\dfrac{-3y_2}{x_2+2}\)
\(y_M-y_N=\dfrac{3y_1}{x_1+2}+\dfrac{3y_2}{x_2+2}\)
\(=3\cdot\dfrac{(kx_1+m)(x_2+2)+(kx_2+m)(x_1+2)}{x_1x_2+2(x_1+x_2)+4}\)
\(=3\cdot\dfrac{2kx_1x_2+2k(x_1+x_2)+m(x_1+x_2)+4m}{x_1x_2+2(x_1+x_2)+4}\)
聯立\(\begin{cases} y=kx+m\\ \dfrac{x^2}{4}+\dfrac{y^2}{3}=1 \end{cases}\)有\((3+4k^2)x^2+8kmx+4m^2-12=0\)
有\(\begin{cases} x_1+x_2=-\dfrac{8km}{3+4k^2}\\ x_1x_2=\dfrac{4m^2-12}{3+4k^2} \end{cases}\)
則
\(S_{\triangle TMN}=\dfrac{1}{2}\left(1-x_T\right)(y_{M}-y_{N})\)
\(=\dfrac{1}{2}\left(1+\dfrac{m}{k}\right)\cdot 3\cdot \dfrac{2kx_1x_2+2k(x_1+x_2)+m(x_1+x_2)+4m}{x_1x_2+2(x_1+x_2)+4}\)
\(=\dfrac{3}{2}\left(1+\dfrac{m}{k}\right)\cdot\dfrac{8km^2-24k-16k^2m-8km^2+12m+16k^2m}{4m^2-12-16km+12+16k^2}\)
\(=\dfrac{3}{2}\left(1+\dfrac{m}{k}\right)\cdot\dfrac{12m-24k}{4m^2-16km+16k^2}\)
\(=\dfrac{3}{2}\left(1+\dfrac{m}{k}\right)\dfrac{3(m-2k)}{(m-2k)^2}\)
\(=\dfrac{9}{2}\left(1+\dfrac{m}{k}\right)\dfrac{1}{m-2k}\)
\(=\dfrac{9}{2}\cdot\dfrac{k+m}{k(m-2k)}\)
因\(m=1\)，則
\(S=\dfrac{9}{2}\cdot\dfrac{k+1}{k(1-2k)}\)
\(\xlongequal[]{k+1=t}\dfrac{9}{2}\cdot\dfrac{t}{(t-1)(3-2t)}\)
\(=\dfrac{9}{2}\cdot\dfrac{t}{-2t^2+5t-3}\)
\(=\dfrac{9}{2}\cdot\dfrac{1}{5-\left(\dfrac{3}{t}+2t\right)}\)
\(\geq \dfrac{9}{2}\left(5+2\sqrt{6}\right)\)