[填空题]
The following diagram shows the parabolic shrsn:e)4a)ro sggcl :5ape of a gateway arch that has a yqt;f4/0l:r k g1r e z4jt-gtlspan of 12 metres and a maximum height of 8 /-yf;0r4lg:et4lzkqr tg 1tj 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$^2$.
