[填空题]
The following diagram shows tvjg ;wr4j(gy o/c52n2xxl)xkhe parabolic shape of a gatewrm40smu h8fu i3.pgg ;ay arch that has a span of 12 metres and a maximum height of 8 me ifs3rhm4pug. mug 80;tres.
The curve has an equation in the form y=$k(x-6)^{2}+8$ .
1. Determine the value of k is $\frac{a}{b}$;a= ,b= .
2. Write down an integral that represents the cross sectional area under the arch shown as OMN.
A=$\int_{0}^{12}\left[-\frac{a}{b}(x-6)^{2}+c\right] \mathrm{d} x$;a= ,b= ,c=
3. Find the cross sectional area under the arch.Evaluating the integral in part (b), we get a= m$^2$.
