简单的非对称问题
已知点\(F_1(-1,0),F_2(1,0)\)，动点\(M\)满足\(|MF_1|+|MF_2|=4\)，动点\(M\)的轨迹记为\(E\)
\((1)\) 求\(E\)方程
\((2)\) 若不垂直于\(x\)轴的直线\(l\)过点\(F_2,\)与\(E\)交于\(C,D\)两点(\(C\)在\(x\)轴上方)，\(A_1,A_2\)分别是\(E\)的左、右顶点，设直线\(A_1C\)的斜率为\(k_1(k_1\neq 0)\)，直线\(A_2D\)的斜率为\(k_2\)，说明\(\dfrac{k_1}{k_2}\)是定值
解
\((1)\) \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\)
\((2)\) 设\(C(x_1,y_1),D(x_2,y_2),A_1(-2,0),A_2(2,0),CD:my+1=x\)
联立\(\begin{cases}my+1=x\\\dfrac{x^2}{4}+\dfrac{y^2}{3}\end{cases}\)
得\(y^2(4+3m^2)+6my-9=0\)
即\(y_1+y_2=-\dfrac{6m}{4+3m^2},y_1y_2=\dfrac{-9}{4+3m^2}\)
\(\dfrac{k_1}{k_2}=\dfrac{\dfrac{y_1}{x_1+2}}{\dfrac{y_2}{x_2-2}}=\dfrac{y_1}{x_1+2}\cdot \dfrac{x_2-2}{y_2}=\dfrac{y_1}{my_1+3}\cdot \dfrac{my_2-1}{y_2}=\dfrac{my_1y_2-y_1}{my_1y_2+3y_2}\)
而\(y_1+y_2=\dfrac{2m}{3}y_1y_2\)代回有
\(\dfrac{\dfrac{3}{2}(y_1+y_2)-y_1}{\dfrac{3}{2}(y_1+y_2)+3y_2}=\dfrac{y_1+3y_2}{3y_1+9y_2}=\dfrac{1}{3}\)