Question

直线y=﹣2与抛物线y=﹣x²交于A,B两点(1)求△AOB的周长与面积

Answer 1

解方程得 交点(-2^(1/2),-2),(2^(1/2),-2)

AB=2^(1/2)-(-2^(1/2)=2*2^(1/2) 2倍根号2

OA=OB=[(2^(1/2))^2+(-2)^2]^(1/2)=6^(1/2)

周长=2*2^(1/2)+2*6^(1/2)

面积=(1/2)*2*2^(1/2)*2=2*2^(1/2)