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