Question Statement

The control towers are located at points $Q(-500,0)$ and $R(500,0)$ on a straight shore where the $x$-axis runs through. At the same moment both towers send a radio signal to a ship out at sea, each travelling at $300\mathrm{m}/\mu \mathrm{s}$. The ship received the signal from $Q$ $3\mu \mathrm{s}$ earlier than the message from $R$. Find the equation of the hyperbola containing the possible locations of the ship.

---

Background and Explanation

This problem applies the geometric definition of a hyperbola: the locus of points where the absolute difference of distances to two fixed points (foci) is constant. Since radio signals travel at constant speed, the time difference translates directly to a distance difference, allowing us to determine the hyperbola's parameters.

---

Solution

Let the ship be at position $S$. According to the definition of a hyperbola with foci at $Q$ and $R$:

$|SR|-|SQ|=2a$

(Note: Since the ship receives the signal from $Q$ earlier, it is closer to $Q$ than to $R$, making $|SR|>|SQ|$.)

Let the ship hear the signal from $R$ after $t$ microseconds. Then it hears the signal from $Q$ after $(t-3)$ microseconds (since the signal from $Q$ arrives $3\mu \mathrm{s}$ earlier).

Since $\text{Distance}=\text{Velocity}\times \text{Time}$:

$|SR|=300\times t\text{ meters}$

$|SQ|=300\times (t-3)\text{ meters}$

Substituting these values into equation (i):

$300t-300(t-3)=2a$

Simplifying: $\begin{array}{rl}300t-300t+900& =2a\\ 2a& =900\\ a& =450\end{array}$

Since the distance between the foci is $2c$:

$\begin{array}{rl}|QR|& =2c\\ \sqrt{(500-(-500){)}^2+(0-0{)}^2}& =2c\\ \sqrt{(1000{)}^2}& =2c\\ 2c& =1000\\ c& =500\end{array}$

For a hyperbola, the relationship between $a$, $b$, and $c$ is: $c^2=a^2+b^2$

Solving for $b^2$: $\begin{array}{rl}{b}^2& =c^2-a^2\\ b^2& =500^2-450^2\\ b^2& =250000-202500\\ b^2& =47500\\ b& =\sqrt{47500}=50\sqrt{19}\end{array}$

The standard equation of a hyperbola centered at the origin with horizontal transverse axis is:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Substituting the values $a=450$ and $b=50\sqrt{19}$:

$\frac{x^2}{(450{)}^2}-\frac{y^2}{(50\sqrt{19}{)}^2}=1$

Or equivalently:

$\frac{x^2}{202500}-\frac{y^2}{47500}=1$

---

Key Formulas or Methods Used

• Definition of hyperbola: $|P{F}_1-P{F}_2|=2a$ (constant difference of distances to foci)
• Distance-speed-time relationship: $d=v\times t$
• Hyperbola parameter relationship: $c^2=a^2+b^2$ (where $2c$ is focal distance, $2a$ is transverse axis)
• Standard equation: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ for horizontal hyperbola centered at origin

---

Summary of Steps

1. Identify the foci: Locate $Q(-500,0)$ and $R(500,0)$, giving $c=500$.
2. Convert time difference to distance: $3\mu \mathrm{s}\times 300\mathrm{m}/\mu \mathrm{s}=900\mathrm{m}=2a$.
3. Find $a$: $2a=900\Rightarrow a=450$.
4. Calculate $b$: Use $b^2=c^2-a^2=500^2-450^2=47500$.
5. Write the equation: Substitute into $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to get $\frac{x^2}{202500}-\frac{y^2}{47500}=1$.