Question

Two stations, located at M(−1.5, 0) and N(1.5, 0) (units are in km), simultaneously send sound signals to a ship, with the signal traveling at the speed of 0.33 m/s. If the signal from N was received by the ship four seconds before the signal it received from M, find the equation of the curve containing the possible location of the ship.From the problem, how far in km are the two stations?
What is the equation of the curve containing the possible location of the ship?

Answer 1

Co-ordinate of M (-1.5,0)

Co-ordinate of N (1.5,0)

Let, Co-ordinate of Ship (x, y)

Now according to the question,

time taken by signal to reach to N is 4 second less than that of M.

i.e. $\frac{SN}{0.33} + 4 =\frac{SM}{0.33}$

where, $SN = \sqrt{(x-1.5)^2 +y^2}$

$SM = \sqrt{(x+1.5)^2 +y^2}$

So equation will be, $4 =\frac{SM}{0.33}-\frac{SN}{0.33} = \frac{SM-SN}{0.33} \implies SM-SN = \frac{4}{3}$

$\sqrt{(x+1.5)^2 +y^2}-\sqrt{(x-1.5)^2 +y^2} =\frac{4}{3}$

Squaring both sides, we get

${(x-1.5)^2 +y^2} +{(x+1.5)^2 +y^2}-2\sqrt{(x-1.5)^2 +y^2}\sqrt{(x+1.5)^2 +y^2} = \frac{16}{9}$

$2x^2+2y^2+4.5-2\sqrt{x^4+2x^2y^2-4.5x^2+y^4+4.5y^2+5.0625} =1.7$

$2x^2+2y^2+4.5-1.7=-2\sqrt{x^4+2x^2y^2-4.5x^2+y^4+4.5y^2+5.0625}$

$2x^2+2y^2+2.8=-2\sqrt{x^4+2x^2y^2-4.5x^2+y^4+4.5y^2+5.0625}$

Solving equation we get,

$(2x^2+2y^2+2.8)^2-(2\sqrt{x^4+2x^2y^2-4.5x^2+y^4+4.5y^2+5.0625} )^2 = 0$

$29.2x^2-6.8y^2-12.41 = 0$ (1)

Equation (1) is the required curve.

M and N are at a distance of 3.0 Km.

Equation (1) gives the location of the all possible ships