 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2012/9/b/iii/solution

For fixed

t

, eliminate

s = x - t

to obtain the projected parabola

y = x - 2 t + (x - t)^{2} .

Along that curve,

x + y + 1 = (s + 1)^{2} \geq 0.

Equality occurs at

s = - 1

, where the point is

(t - 1, - t)

. Also

d y / d x = y_{s} / x_{s} = 1 + 2 s = - 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 \geq 0

, the attained tangency points form only the half-line

x \geq - 1

y = - x - 1

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