A regular hexagon of side length 1 is placed vertically along one of its edges on a flat surface. A particle is fired from the surface and passes through four of the vertices of the hexagon, as shown below, tracing the arc of a parabola. If the particle starts at A and ends at B, then find the length AB.
+128408 Look at the graph here :
Let O = (0, 0)
By the Law of Cosines we can find the height, h, of point F above the x axis
h^2 = 1^2 + 1^2 - 2(1)(1)cos120
h^2 = 2 - 2(1/2)
h^2 = 2 - 1
h = sqrt(3).....so F = (.5, sqrt(3) )
And D = (1, sqrt(3)/2 )
And let the vertex of the parabola be (0, k)
Since the parabola passes through D and F we have that
sqrt(3) = a( .5)^2 + k
sqrt(3)/2 = a(1 )^2 + k subtract these and we have
sqrt(3)/2 = a (-3/4)
-4sqrt(3)/6 = a
-2sqrt(3)/3 = a
-2/sqrt(3) = a
So....we can find k as
sqrt (3) = -2/sqrt(3)(.5)^2 + k
sqrt(3) + 2/ [4sqrt(3) ] = k
[12 + 2 ] / [ 4sqrt(3) ] = k
7/[2sqrt(3) ] = k
To find the positive x coordinate of "B" we have
0 = -2/[sqrt(3)] (x)^2 + 7/ [2sqrt(3)]
-7/ [ 2sqrt(3) ] = - 2/ [ sqrt(3) ] (x)^2
7/4 = x^2
sqrt(7)/2 = x
So....by symmetry, AB = 2 *sqrt(7) / 2 = sqrt (7)
