(i) We place the parabola so that the left side of the arch passes through `(0, 0)` and the right side will pass through `(2, 0)`, since the arch is 2 m wide at the bottom. The vertex of the arch is at `(1, 3)`.

General form of a parabola: `y = ax^2+ bx + c`

At `x = 0`, `y = 0` and on substituting, we get `0 = 0 + 0 + c`. So `c = 0`.

At `x = 1`, `y = 3` and on substituting, we get `3 = a + b`.

At `x = 2`, `y = 0` and on substituting, we get `0 = 4a + 2b`.

This gives the simultaneous equations

`a + b = 3`

`2a + b = 0`

Subtracting the 1st line from the 2nd gives `a = -3`.

And so `b = 6`.

So the required parabola is `y = -3x^2+ 6x`, with `x` in metres.

Parabolic arch0.20.40.60.811.21.41.61.82123xf(x)Open image in a new page

Parabolic arch, 2 m wide and 3 m high..

The process of finding the equation is called modeling. It is a very important skill in science and engineering.

(ii) Now for the area:

`int_0^2(-3x^2+6x)dx =[-x^3+3x^2]_0^2`

`=[-(2^3)+3(2)^2]-[0+0]`

`=[-8+12]`

`=4\ "m"^2`

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