The curve of the cross section is of the form

y = ax^2

Put in one of the given values and solve for "a".

40 = a(150^2)

40/150^2 = a

40/22500 = a = 1/5625

The curve is

The depth is given as 40', and the y value 150' from the origin is 40. We assume, then, that the origin is the vertex of the curve. Thus, y = a(x-h)^2 + k reduces to y = ax^2, as (h, k) = (0, 0).

y = ax^2

Put in one of the given values and solve for "a".

40 = a(150^2)

40/150^2 = a

40/22500 = a = 1/5625

The curve is

**y = (x^2)/5625**

The depth is given as 40', and the y value 150' from the origin is 40. We assume, then, that the origin is the vertex of the curve. Thus, y = a(x-h)^2 + k reduces to y = ax^2, as (h, k) = (0, 0).