Question

椭圆方程x²/2+y²=1，直线l过点P(0，2)且与椭圆相交于AB两点，当△AOB面积取得最大

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Answer 1

L:y=kx+2
x^2/2+y^2=1
x^2+2*(kx+2)^2=2
(1+2k^2)x^2+8kx+6=0
xA+xB=-8k/(1+2k^2),xA*xB=6/(1+2k^2)
(xA-xB)^2=(xA+xB)^2-4xA*xB=(16k^2-24)/(1+2k^2)^2
AB^2=(1+k^2)*(xA-xB)^2=(1+k^2)*(16k^2-24)/(1+2k^2)^2
y=kx+2
kx-y+2=0
d^2=|0-0+2|^2/(1+k^2)=4/(1+k^2)
s^2=(1/4)*AB^2*d^2=(1/4)*[(1+k^2)*(16k^2-24)/(1+2k^2)^2]*[4/(1+k^2)]
s^2≤1/2
s^2(max)=1/2
k=±√3.5
L:y=2±(√3.5)x