We've all heard of the special property that parabolas have; namely, when a beam in the interior of a parabola and parallel to the parabola's axis of symmetry hits the parabola it then is such reflected that it passes through the focal point of the parabola. Now the proof of this is not too difficult and is a good exercise in analytic geometry (and maybe a bit of calculus). But here's a question: Is parabola unique in this regard among all axially symmetric curves? In other words, if a curve has the reflection property noted above, then is that curve necessarily a parabola?

say (0,x_0) is the focal, by analytic geometry (and maybe a bit of trigonometry) you see that a solution should verify something likeand if you fix an initial condition outside the origin then this should have a unique solution on a neighbourhood of it.

just to fill in a little more details:set the focal point at (0,0). the light beam follows x=x0, get reflected at point A on the curve: (x0,y0), then arrives at the origin O: (0,0). say the normal line of the curve at A (the angle bisector) intersects y-axis at B. easy to see OAB is an isosceles triangle, hence B is at (0,+/-sqrt[x0^2+y0^2]), which gives the slope of AB:(y0 +/- sqrt[x0^2+y0^2])/x0.the tangent line at A has a slope of dy/dx at x=x0. so you have an ODE:(or take x0=0 in arkol's eq.)thusintegrating both sides gives +/-y=sqrt(x^2+y^2)+C (essentially the parabola eq. in polar coordinates). exclude the trivial case C=0, this is a one-parameter family of parabolas: y=(x^2-C^2)/(2C). note two parabola have C's with same sign don't intersect, so one cannot construct a curve that is parabola piecewise.