第2个回答 2012-01-07
⑴∵e=a/c=√2/2,
∴b=c=√2/2a,即a^2=2b^2=2c^2,
∴椭圆方程可化为x^2/2b^2+y^2/b^2=1,
将点A坐标(2,1)代入,得:
4/2b^2+1/b^2=1,即3/b^2=1,∴b^2=3,∴a^2=2b^2=6
∴椭圆方程为x^2/6+y^2/3=1
⑵方法是设而不求,即设M、N坐标分别为(x1,y1)、(x2,y2)
再设出直线l的方程(斜率为k),与椭圆方程联立,再代入,即可求出!
sorry啊,过程有点难打上来,所以就教一下方法,sorry!
第3个回答 2011-12-30
1 4/a²+1/b²=1,a²=b²+c²,c/a=√2/2.,
a=6, b=3.
x²/6+y²/3=1.
2 设ly=k(x-3),
y=k(x-3),x²/6+y²/3=1
(1+2k²)x²-12k²x+18k²-6=0.
设M,N为(x1,y1),(x2,y2),
x1+x2=12k²/1+2k², x1x2=18k²-6/1+2k²,
y1=k(x1-3),y2=k(x2-3).
kAM+kAN= y1-1/x1-2+y2-1/x2-2
= (kx1-3k-1)(x2-2)+(kx2-3k-1)(x1-2)/(x1-2)(x2-2)= 2kx1x2-(5k+1)(x1+x2)+12k+4/x1x2-2(x1+x2)+4
= 2k(18k²-6)-(5k+1)•12k²+(12k+4)(1+2k²)/18k²-6-24k²+4(1+2k²)= -4k²+4/2k²-2=-2.
kAM+kAN为定值-2.本回答被提问者采纳