把y=x+m①代入x^2+y^2/4=1,整理得
5x^2+2mx+m^2-4=0,
△=4m^2-20(m^2-4)=16(5-m^2)
設(shè)A(x1,y1),B(x2,y2),P(x,y),則
x1,2=[-m土2√(5-m^2)]/5,
由AP/PB=2得(x-x1,y-y1)=2(x2-x,y2-y),
∴x=(x1+2x2)/3=(-3/5)m土(2/5)√（5-m^2),
由①,m=y-x,代入上式化簡(jiǎn)得
(2x+3y)^2=4[5-(y-x)^2],
即8x^2+4xy+13y^2=20,為所求.