Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 9 / b / iii

For fixed $t$, eliminate $s=x-t$ to obtain the projected parabolaAlong that curve,Equality occurs at $s=-1$, where the point is $(t-1,-t)$. Also $dy/dx=y_s/x_s=1+2s=-1$ there, so the tangent line has the same slope as the line $x+y=-1$ and passes through a point of that line. Thus every projected characteristic is tangent to that line, with no crossing of it. Because $t\ge 0$, the attained tangency points form only the half-line $x\ge -1$, $y=-x-1$.