已知椭圆C:x2⼀a2+y2⼀b2=1(a>b>0)的离心率为√2⼀2,点(2,√2)在C上

2024-11-19 20:22:09
推荐回答(1个)
回答1:

心率为√2/2即a^2=2c^2;所以:b^2=a^2-c^2=2c^2-C^2=C^2
椭圆C:x2/a2+y2/b2=1(a>b>o)经过点A(2,1),那么4/2c^2+1/c^2=1;解得:c^2=3
所以:a^2=6,b^2=3椭圆为:x^2/6+y^2/3=1
(2)设M(x1,y1),N(x2,y2);因为点B(3,0在椭圆外,所以直线l的斜率一定存在;
设直线l的方程为:y=k(x-3)代入椭圆方程中得:(1+2k^2)x^2-12k^2x+18k^2-6=0
由判别式>0得:-1x1+x2=12k^2/(1+2k^2);x1x2=(18k^2-6)/(1+2k^2)
|BM|×|BN|=√(1+k^2)|x1-3|√(1+k^2)|x2-3|=(1+k^2)|(x1-3)(x2-3)|=(1=k^2)|x1x2-3(x1+x2)+9|
=3(1+k^2)/(1+2k^2)=(3/2)[1+1/(1+2k^2)]
由-1(3)kAM=(y1-1)/(x1-2)=(kx1-3k-1)/(x1-2);kAN=(y2-1)/(x2-1)=(kx2-3k-1)/(x2-2)
所以kAM+KAN=(kx1-3k-1)/(x1-2)+(kx2-3k-1)/(x2-2)化简即可