Three receivers \(P_1\text{,}\) \(P_2\text{,}\) and \(P_3\) are deployed in the landscape. The figure captures the distances we know. Emma’s tourist navigation device sends a signal to all three receivers. The signal arrives at receivers \(P_1\) and \(P_3\) at the same time, and at receiver \(P_2\) it arrives \(80\) microseconds later. Where is Emma located? Assume the signal travels \(300{,}000\) km per second, and determine the position in a suitably chosen coordinate system.
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Solution.
We first translate the two timing facts into geometry. The signal reaches \(P_1\) and \(P_3\) simultaneously, so Emma is equidistant from them: he lies on the perpendicular bisector of the segment \(P_1P_3\text{.}\) The signal reaches \(P_2\) \(80\) microseconds after \(P_1\text{,}\) so Emma is
\begin{equation*}
(300{,}000\ \text{km/s})\,(80\times 10^{-6}\ \text{s}) = 24\ \text{km}
\end{equation*}
farther from \(P_2\) than from \(P_1\text{.}\) He therefore also lies on the branch nearer \(P_1\) of the hyperbola \(h\) with foci \(P_1\) and \(P_2\) whose constant distance difference is \(24\) km.
Now we choose coordinates so that \(h\) has the simplest possible equation. Place the origin \(O\) at the midpoint of \(P_1P_2\text{,}\) point the positive \(x\)-axis along \(OP_1\text{,}\) and choose the positive \(y\)-axis so that \(P_3\) has positive second coordinate. Since every given length is a multiple of \(12\text{,}\) let one unit on each axis be \(12\) km. Then
\begin{equation*}
P_1(2,0), \qquad P_2(-2,0), \qquad P_3(3,3).
\end{equation*}
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Let \(A\) denote Emma’s unknown position, and let \(o\) be the perpendicular bisector of \(P_1P_3\text{.}\) The line \(o\) passes through the midpoint \(S\left(\tfrac52, \tfrac32\right)\) of the segment, and since \(P_3 - P_1 = (1,3)\text{,}\) the direction \(\vec u_o = (3,-1)\) is perpendicular to the segment. Parametrically,
\begin{equation*}
o:\quad x = \tfrac52 + 3t, \qquad y = \tfrac32 - t, \qquad t \in \mathbb{R}.
\end{equation*}
For the hyperbola, the foci are \(P_1(2,0)\) and \(P_2(-2,0)\text{,}\) so the center is \(O\) and \(c = |OP_1| = 2\text{.}\) The constant difference \(|AP_2| - |AP_1| = 24\ \text{km} = 2\) units equals \(2a\text{,}\) so \(a = 1\text{,}\) and then \(b^2 = c^2 - a^2 = 4 - 1 = 3\text{.}\) Hence
\begin{equation*}
h:\quad x^2 - \frac{y^2}{3} = 1,
\end{equation*}
and since \(A\) is closer to \(P_1\text{,}\) it lies on the right branch: \(x_A \gt 0\text{.}\)
It remains to intersect \(o\) with \(h\text{.}\) Substituting the parametric equations into \(3x^2 - y^2 = 3\text{:}\)
\begin{align*}
3\left(\tfrac52 + 3t\right)^2 - \left(\tfrac32 - t\right)^2 \amp= 3\\
3\left(\tfrac{25}{4} + 15t + 9t^2\right) - \left(\tfrac94 - 3t + t^2\right) \amp= 3\\
26t^2 + 48t + \tfrac{27}{2} \amp= 0\\
52t^2 + 96t + 27 \amp= 0.
\end{align*}
The discriminant is \(96^2 - 4\cdot 52\cdot 27 = 9216 - 5616 = 3600 = 60^2\text{,}\) so
\begin{equation*}
t = \frac{-96 \pm 60}{104}, \qquad t_1 = -\frac{9}{26}, \quad t_2 = -\frac32.
\end{equation*}
Substituting \(t_1\) into the parametric equations gives
\begin{equation*}
x_1 = \tfrac52 + 3\left(-\tfrac{9}{26}\right) = \tfrac{19}{13}, \qquad
y_1 = \tfrac32 - \left(-\tfrac{9}{26}\right) = \tfrac{24}{13},
\end{equation*}
that is, \(A_1\left(\tfrac{19}{13}, \tfrac{24}{13}\right)\text{.}\) Substituting \(t_2\) gives \(x_2 = \tfrac52 - \tfrac92 = -2\) and \(y_2 = \tfrac32 + \tfrac32 = 3\text{,}\) that is, \(A_2(-2,3)\text{.}\)
The point \(A_2\) fails the branch condition \(x_A \gt 0\text{,}\) and the physics says why: \(A_2\) lies on the left branch, \(24\) km closer to \(P_2\) than to \(P_1\text{,}\) so a signal sent from \(A_2\) would reach \(P_2\) earlier than \(P_1\)—the opposite of what was measured. Emma’s position is therefore
\begin{equation*}
A\left(\tfrac{19}{13},\ \tfrac{24}{13}\right),
\end{equation*}
about \((17.5,\ 22.2)\) in kilometers from \(O\text{.}\) As a check, the distances come out exactly: \(|AP_1| = |AP_3| = \tfrac{25}{13}\) units and \(|AP_2| = \tfrac{51}{13}\) units, so \(|AP_2| - |AP_1| = 2\) units \(= 24\) km, as required. In kilometers, Emma is \(\tfrac{300}{13} \approx 23.1\) km from \(P_1\) and \(P_3\text{.}\)
