Question

设F1 F2是椭圆x^2/8+y^2/4=1的左右焦点,过F2的直线交椭圆于A B两点,且|AB|=3根号2,求S△AF

Answer 1

a=8,b=4,c=2F2（2,0）A(x1,y1）,B（x2,y2）则设直线y=kx-2k代入x^2/8+y^2/4=1x^2+2(kx-2k)^2=8(2k^2+1)x^2-8k^2x+8k^2-8=0x1+x2=8k^2/(2k^2+1),x1x2=(8k^2-8)/(2k^2+1)y1+y2=k(x1+x2)-4k=-4k/(2k^2+1),y1y2=k^2(x1x2-2(x1+x2)+4)=-4/(2k^2+1)则（x1-x2）+(y1-y2）=18解得k=（1+2√34)/18代入y1-y2=√（48k^2+16)/(2k^2+1)=S△AF1B=4*|y1-y2|/2=2|y1-y2|= 我算的直线斜率很诡异,就没敢往里代入,但是思路是这样的,你做做看哪里算错了