let us choose a coordinate system so that the X axis
passes through the stations M,N
and the Y axis through the middle of the segmentMN
let MN=2c then
M(−c,0);N(−c;0)
S(x,y)−coordinates of points satisfying the condition:
SM−SN=2a
2a−this is the distance traveled by the sound for the difference in time
of signal fixation by the receiving stationsM,N
SM=(x+c)2+y2
SN=(x−c)2+y2
SM>SN
(x+c)2+y2>(x−c)2+y2
(x+c)2+y2>(x−c)2+y2
(x+c)2>(x−c)2
x2+2xc+c2>x2−2xc+c2
4xc>0
x>0(1)
SM−SN=2a
(x+c)2+y2−(x−c)2+y2=2a(2)
(x+c)2+y2=2a+(x−c)2+y2
(x+c)2+y2=4a2+4a(x−c)2+y2+(x−c)2+y2
−4a2+4xc=4a(x−c)2+y2
−a2+xc=a(x−c)2+y2
a4−4axc+x2c2=a2((x−c)2+y2)
x2(c2−a2)−a2y2=a2(c2−a2)
a2x2−c2−a2y2=1(3)
Combining formula (1) and (2), we got the total solution:
S(x,y)
a2x2−c2−a2y2=1 forx>0
This solution can be obtained in a shorter way using the definition of hyperbola
let’s apply specific data from the task:
M(1.5,0);N(1.5,0)
c=0
M(0,0);N(0,0)
2a=0.33∗4=1.32
According to the formula (2)
(x+c)2+y2−(x−c)2+y2=2a
(x+0)2+y2−(x−0)2+y2=1.32
0=1.32 its false
For the given initial conditions, the problem has no solution
Answer:For the given initial conditions, the problem has no solution
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