[填空题]
The following diagram show0l v.qovu:d4 o.wwk/ ls the parabolic shap 4lz8+rxfb3 qhftu 8nh1 9fh.ve of a gateway arch that has a span of 12 metres and a maxiv.z8hu nhf9 lf8 hxrb4q1f+3 tmum 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$^2$.
