椭圆x^2/a^2+y^2/b^2=1(a>0,b>0)的顶点为A1,A2,B1,B2,焦点为F1,F2,AB为根号7,平行四边形_数学解答

椭圆x^2/a^2+y^2/b^2=1(a>0,b>0)的顶点为A1,A2,B1,B2,焦点为F1,F2,AB为根号7,平行四边形A1B1A2B2=平行四边
形A1B1A2B2面积的2倍,求椭圆C的方程?

人气：283 ℃　时间：2020-04-10 17:52:46

解答

S A1B1A2B2=0.5A1A2*B1B2=0.5*2a*2bS B1F1B2F2=0.5F1F2*B1B2=0.5*2c*2bS A1B1A2B2=2S B1F1B2F2 ==>0.5*2a*2b=2c*2b ==>a=2c .(1)|A1BA1|^2=a^2+b^2=7 .(2)a^2=b^2+c^2 .(3)(1)(2)(3)联立解得a^2=4b^2=3c^2=1综上可得...