[填空题]
The following diagram shows the parabolic shapt94h k 5ttal -p.jiw. jkk+vo+d*n ak( r/rn9ne of a gateway zpfr1644d q) :tqw nvlb(go9j+ wd b8xarch that has a span of 12 metres and a m 46jlg4 o:bz(1pdw+ftnwq)r8 vd9xb q aximum height of 8 metres.
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$^2$.
