如图,圆o的半径为6,线段AB与圆o相交于点C、D,AC=4,AC=4,∠BOD=∠A,OB与圆o相交于点E,设OA=x,CD=y,求Y关于x的函数解析式,并求出自变量取值范围

Question

如图,圆o的半径为6,线段AB与圆o相交于点C、D,AC=4,AC=4,∠BOD=∠A,OB与圆o相交于点E,设OA=x,CD=y,求Y关于x的函数解析式,并求出自变量取值范围

Answer 1

AC=4,AC=4?
连接OC,则∠AOC=∠ABO,AC/CO=OD/DB=AO/BO,DB=OD/(AC/CO)=6/(4/6)=9,AO/BO=4/6
BO=6AO/4=3X/2
OD/AO=BO/AB,6/X=(3X/2)/(4+Y+9)=3X/(26+2Y),26+2Y=3X^2/6=X^2/2,Y=X^2/4-13
根号52