连接aq
cp=bc-bp=4-x
pq=√(cp^2-cq^2)=√[(4-x)^2-y^2]
ap=√(ab^2-bp^2)=√(16-x^2)
dq=dc-cq=4-y
aq=√(ad^2-dq^2)=√[16-(4-y)^2]
因为,ap垂直pq
则在直角三角形aqp中
aq^2=ap^2+pq^2
16-(4-y)^2=16-x^2+(4-x)^2-y^2
16-16+8y-y^2=16-x^2+16-8x+x^2-y^2
y=-x+2