非对称韦达定理
已知椭圆\(E\)的左焦点为\((-2\sqrt{2},0)\),长轴长为\(8\)
\((1)\) 求椭圆\(E\)的标准方程
\((2)\) 记\(E\)的左右定点分别为\(A,B\),过点\(C(2,0)\)的直线\(l\)与\(E\)交于\(M,N\)两点(\(M,N\)均不与\(A,B\)重合),直线\(MA,NB\)交于点\(P\),证明\(P\)在一条定直线上,并求出此定直线.
\((1)\) \(\dfrac{x^2}{16}+\dfrac{y^2}{4}=1\)
\((2)\) 设\(M(x_1,y_1),N(x_2,y_2),MN;my+2=x\)
则\(MA:y=\dfrac{y_1}{x_1+4}(x+4),NB:y=\dfrac{y_2}{x_2-4}(x-4)\)
联立\(MA,NB\)有
\(\dfrac{x+4}{x-4}=\dfrac{y_2(x_1+4)}{y_1(x_2-4)}=\dfrac{my_1y_2+6y_2}{my_1y_2-2y_1}\)
联立\(MN,E\)有
\(y^2(4+m^2)+4my-12=0\)
即\(y_1+y_2=-\dfrac{-4m}{4+m^2},y_1y_2=-\dfrac{12}{4+m^2}\)
则有\(3(y_1+y_2)=my_1y_2\)
从而带回有\(\dfrac{x+4}{x-4}=\dfrac{3y_1+3y_2+6y_2}{3y_1+3y_2-2y_1}=\dfrac{9y_2+3y_1}{3y_2+y_1}=3\)