y=kx+2
X^2/2+y^2=1
联解,化简,得
(2k^2+1)x^2+8kx+6=0
AB长=√k^2+1√(16k^2-24)/(2k^2+1)
再求(0,0)到y-kx-2的距离
d=2/√k^2+1
所以S=√(16k^2-24)/(√k^2+1)
y=kx+2
X^2/2+y^2=1
联解,化简,得
(2k^2+1)x^2+8kx+6=0
AB长=√k^2+1√(16k^2-24)/(2k^2+1)
再求(0,0)到y-kx-2的距离
d=2/√k^2+1
所以S=√(16k^2-24)/(√k^2+1)