Question

如图,在平面直角坐标系中,AB垂直于x轴于点B,AB=3,tan角AOB=四分之三,将三角形OAB绕原点O逆时针旋转90

Answer 1

没有图象,假设A在第二象限.

⑴tan∠AOB=AB/OB=3/4,AB=3,∴OB=4,

∴B(-4,0),B1(0,-4),A2(3,0),

设解析式为Y=a(X+4)(X-3),-4=-12a,a=1/3,

∴Y=1/3X^2+1/3X-4,

⑵直线BB1解析式为Y=Y=-X-4,设P(m,1/3m^2+1/3m-4),

过P作PR⊥X轴于R,交BB1于Q,则Q(m,-m-4),

PQ=-m-4-(1/3m^2+1/3m-4)=-1/3m^2-4/3m,

SΔPBB1=SΔPQB+SΔPQB1=1/2PQ*BR+1/2PQ*OR

=1/2PQ*OB=2PQ=-2/3(m^2+2m)=-2/3(m+1)^2+8/3,

∴当m=-1时,SPBB1最大=8/3,

这时P(-1,-4),

⑶以BB1为底的ΔQBB1,SΔQBB1=1/2BB1*√2/2=2