Question

如图点P在反比例函数y=4/x(x>0）上一点,PM⊥x轴于M点,交y=1/x于A点,PH⊥y轴于N点,交y=1/x于B

Answer 1

(1)

OM*AM=1

OD*DB=1

∴S(OMAC)=S(ODBN)=1

S(AMDH)=S(OMAC)-S(ODHC)

=S(ODBN)-S(ODHC)

=S(BNCH)

(2)

OA=OB

OA²=OB²

OM²+MA²=OD²+DB²

(OM+MA)²=OM²+MA²+2OM*MA

=OD²+DB²+2OD*DB

=(OD+DB)²

OM+MA=OD+DB

同理

(OM-MA)²=OM²+MA²-2OM*MA

=OD²+DB²-2OD*DB

=(OD-DB)²

OM-MA=OD-DB

因此OM=DB

MA=OD

PA=DB-AM

=OM-OD

=PB

(3)

设P(x0,y0)

x0y0=4

H(1/y0,1/x0)

S(HDOC)=1/(x0y0)=1/4

(4)

k(CD)=-y0/x0

A(x0,1/x0)

B(y0,1/y0)

k(AB)=(1/y0-1/x0)/(y0-x0)

=-1/(x0y0)

=-1/4是定值

所以

当P(1,4)时AB∥CD

P取其它点时AB,CD相交.

如仍有疑惑,欢迎追问.祝：