题目

已知离心率为1/2的椭圆C:x²/a²+y²/b²=1(a>b>0)经过点P(1.3/2),A1是椭圆C的右顶点.①求椭圆C的方程.②若直线l:x=my+2/7与椭圆C相交于A,B俩点,求证:向量A4A×向量A1B=0

提问时间：2020-12-13

答案

①c/a=1/2,∴a=2c,b^2=3c^2,
∴椭圆x^2/(4c^2)+y^2/(3c^2)=1过点(1,3/2),
∴1/c^2=1,c^2=1,
∴椭圆的方程是x^2/4+y^2/3=1.①
②A1(2,0),把x=my+2/7②代入①*12,得3[m^2y^2+(4/7)my+4/49]+4y^2=12,
整理得(3m^2+4)y^2+(12/7)my-576/49=0,
设A(x1,y1),B(x2,y2),则y1+y2=-12m/[7(3m^2+4)],y1y2=-576/[49(3m^2+4)],
∴向量A1A*A1B=(x1-2,y1)*(x2-2,y2)=(x1-2)(x2-2)+y1y2
=(my1-12/7)(my2-12/7)+y1y2(由②)
=(m^2+1)y1y2-(12/7)m(y1+y2)+144/49,
=[-576(m^2+1)+144m^2+144(3m^2+4)]/[49(3m^2+4)]
=0.