如图,A(0,4)B(3,0)C(4,2),且反比例函数图象经过点C
1个回答

(1) A(0,4) B(3,0) C(4.2) D(x1,y1)

|AB|=√((0-3)^2+(4-0)^2)=5

kAB=(4-0)/(0-3)=-4/3

平行四边形:|CD|=|AB| kCD=kAB

√((x1-4)^2+(y1-2)^2)=5

(x1-4)^2+(y1-2)^2=25.(1)

(y1-2)/(x1-4)=-4/3

y1-2=-4/3(x1-4).(2)

(2)代入(1):(x1-4)^2+(-4/3(x1-4))^2=25

(x1-4)^2(1+16/9)=25

(x1-4)^2=9

x1-4=±3

x1=4+3=7

y1-2=-4/3(7-4)

y1=-2

或者x1=4-3=1

y1-2=-4/3(1-4)

y1=6

D(7,-2)或者D(1,6)

(2) C(4,2)在y=k/x上

2=k/4

k=8

y=8/x

设Q(q,8/q)到AB的距离为d (q>0)

AB方程:(y-0)/(x-3)=(4-0)/(0-3)

4x+3y+12=0

d=|4q+3*8/q+12|/√(4^2+3^2)

=4|q+6/q+3|/5

SΔABQ=1/2|AB|*4|q+6/q+3|/5

=2/5*5|q+6/q+3|

=2|q+6/q+3|

q+6/q>=2√q*√(6/q)=2√6

当q=6/q时,q+6/q有最小值:2√6

SΔABQmin=2|2√6+3|=4√6+6

(2) 解法2:

C(4,2)在y=k/x上

2=k/4

k=8

y=8/x

当y=8/x上某点的切线与AB平行时,三角形ABQ的面积最小.

y'=-8/x^2=kAB=-4/3

x^2=6

x=±√6(∵在第一象限,∴x=√6)

y=8/√6=4√6/3

Q(√6,4√6/3)

AB方程:(y-0)/(x-3)=(4-0)/(0-3)

4x+3y+12=0

Q到AB的距离:d=|4*√6+3*4√6/3+12|/√(4^2+3^2)

=(8√6+12)/5

SΔABQmin=1/2*|AB|*d

=1/2*5*(8√6+12)/5

=4√6+6