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Solve Lesson 5 Page 56 Math Study Topic 10 – Kite>


Topic

Along the coast, people set up long-range radio navigation systems to transmit signals to aircraft or ships operating at sea. In that system there are two radio stations located at location A and location B respectively, the distance AB = 650 km (Figure 18). Suppose there is a ship moving on the sea with a hyperbolic orbit that has A and B as the two foci.

While in position P, the receiver on board the ship converts the time difference of receiving signals from A and B into a distance signal \(\left| {PA – PB} \right|\). Assume that the time it takes the ship to receive a signal from B before it receives a signal from A is 0.0012 s. The signal travel speed is \({3.10^8}\) m/s.

a) Prepare a hyperbolic equation describing the trajectory of the ship’s motion

b) Show that at any time in the moving trajectory, the time it takes for the ship to receive a signal from B before it receives a signal from A is always 0.0012 s.

Detailed explanation

Since the time it took the ship to receive a signal from B before it received a signal from A was 0.0012 s, at that time \(PA – PB = \left( {{{3.10}^8}} \right) .0.0012 = 360,000\left( m \right) = 360\left( {km} \right)\)

Since the eyebrow moves with a hyperbolic trajectory, A and B are the two foci, so\(\left| {PA – PB} \right| = 360\left( {km} \right)\) for all positions of P

Choose the coordinate system so that the origin coincides with the midpoint of AB and the Ox axis coincides with AB, the unit on both axes is km, then this hyperbola has the form \(\frac{{{x^2}}}{{ {a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1\) where \(a > 0,b > 0\)

Since \(\left| {PA – PB} \right| = 360\) \(2a = 360 \Rightarrow a = 180\)

According to the problem, \(AB = 650 \Rightarrow 2c = 650 \Rightarrow c = 325\)

We have: \({b^2} = {c^2} – {a^2} = {325^2} – {180^2} = 73225\)

So the hyperbolic equation takes the form \(\frac{{{x^2}}}{{32400}} – \frac{{{y^2}}}{{73225}} = 1\)

b) Since the ship moves only in the right arm of the Oy axis of the hyperbola, we PB < PA for all positions of P. Hence, the train always receives a signal from B before it receives a signal from A.

Let \({t_1}\) be the time it takes for the train to receive a signal from A, \({t_2}\) is the time it takes for the ship to receive a signal from B, then \({t_1} = \frac{{PA) }}{v},{t_2} = \frac{{PB}}{v}\) where v is the speed of the signal.

Then we have:

\({t_1} – {t_2} = \frac{{PA – PB}}{v} = \frac{{360000}}{{{{3.10}^8}}} = 0.0012\left( s \) right)\)

So the time it takes the ship to receive a signal from B before it receives a signal from A is always 0.0012 s.



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