x=-2y+2
代入
b^2(-2y+2)^2+a^2y^2=a^2b^2
(4b^2+a^2)y^2-8b^2y+4b^2-a^2b^2=0
y1+y2=8b^2/(4b^2+a^2)
AB中點縱坐標是1/2
所以y1+y2=2*1/2
所以8b^2/(4b^2+a^2)=1
a^2=4b^2
代入(4b^2+a^2)y^2-8b^2y+4b^2-a^2b^2=0
8b^2y^2-8b^2y+4b^2-4b^4=0
2y^2-2y+1-b^2=0
y1+y2=1,y1y2=(1-b^2)/2
所以(y1-y2)^2=(y1+y2)^2-4y1y2=1-2(1-b^2)=-1+2b^2
x=-2y+2
(x1-x2)^2=[(-2y1+2)-(-2y2+2)]^2=4(y1-y2)^2
所以AB^2=(x1-x2)^2+(y1-y2)^2=(√5)^2
所以5(y1-y2)^2=5
(y1-y2)^2=-1+2b^2=1
b^2=1
a^2=4
x^2/4+y^2=1