# Translations:Raypath for velocity linear with depth/18/en

Note that the vertical component ${\displaystyle -v_{0}/a}$ does not depend on the angle ${\displaystyle {\theta }_{0}}$. As a result, the family of rays with the given velocity function is made up of the origin and that have their centers on the line ${\displaystyle y{\rm {=}}-v_{0}/a.}$. The circular raypath starts at the origin (which is the shot point). The tangent to the circle at the shot point makes an angle ${\displaystyle {\theta }_{0}}$ with the positive y-axis.
Often, we will know only that the ray passes through a certain point, say, ${\displaystyle B{\rm {=}}\left(x{\rm {,\ }}y\right)}$, and we must determine the equation of the ray. This problem can be solved as described here and shown in Figure 12. Draw line OB. The midpoint of this line is ${\displaystyle A{\rm {=}}\ {\rm {(}}x/{\rm {2}},{\textit {y/2}}{\rm {)}}}$. The center of the raypath circle must lie on the perpendicular bisector of the line OB. The center of the raypath circle also must lie on the horizontal line through the point ${\displaystyle \left({\rm {0,\ }}-v_{0}/a\right)}$. The intersection of these two lines of determination gives the center E ${\displaystyle E{\rm {=}}\left(x_{E}{\rm {,\ }}-v_{0}/a\right)}$ of the raypath circle. From the geometry, we see that