Question

已知直线AC的解析式为y=1/2x+2,其分别交x轴,y轴于A,C两点,与反比例函数y=k/x(x＞0）交于点P

Answer 1

AC方程：y=x/2+2

A：y=0 x/2+2=0 x=-4 A(-4,0)

C：x=0 y=0/2+2=2 C(0,2)

SΔAOC=1/2*|-4|*|2|=4

设P(p,p/2+2) (p>0)

SΔPAB=9

1/2*|p-(-4)|*(p/2+2)=9

(p+4)(p+4)/2=18

(p+4)²=36

p+4=±6

p=-4±6

∵p>0

∴p=-4+6=2

p/2+2=2/2+2=3

P(2,3)

B(2,0)

y=k/x (x>0)

3=k/2

k=6

y=6/x (x>0)

设R(r,6/r) (r>2)

M(r,r/2+2)

T(r,0)

SΔBTM=1/2*(r-2)*(r/2+2)

=(r-2)(r+4)/4

=(r²+2r-8)/4

SΔBTM=SΔAOC

(r²+2r-8)/4=4

r²+2r-8=16

r²+2r-24=0

(r+6)(r-4)=0

r=-6