3.19. To construct the geometric locus of points on the plane, the ratio of whose distances to the point \( F(4.5; 0) \) and to the line \( x=0.5 \) is three.
Solution. Let's make a schematic

![](https://cdn.mathpix.com/cropped/2024_05_22_f5a9bb7960904462604bg-131.jpg?height=308&width=312&top_left_y=1117&top_left_x=146)

Fig. 3.13 diagram (Fig. 3.13).

Let the coordinates of the current point of the desired curve be \( M(x ; y) \). Then the distance from the point \( M(x ; y) \) to the line \( x=0.5 \) is \( d_{1}=|x-0.5| \), and the distance from the point \( M(x ; y) \) to the point \( F(4.5 ; 0) \) is

\[
r_{1}=\sqrt{(x-4.5)^{2}+(y-0)^{2}}.
\]

From the condition of the problem, it follows that \( r_{1}=3 d_{1} \), hence

\[
\sqrt{(x-4.5)^{2}+(y-0)^{2}}=3|x-0.5| \text{. }
\]

After squaring both sides and combining like terms, we obtain the canonical equation of the hyperbola: \( \frac{x^{2}}{2.25} - \frac{y^{2}}{18} = 1 \).