4.
\((\)一\()\)选修\(4-4\):坐标系与参数方程
已知曲线\(C\)的极坐标方程是\(1+3{si}{{{n}}^{2}}\theta =\dfrac{4}{{{\rho }^{2}}}\),以极点为平面直角坐标系的原点,极轴为\(x\)轴的正半轴,建立平面直角坐标系,直线\(l\)的参数方程是\({ }\!\!\{\!\!{ }\begin{matrix} x=10+2t \\ y=-2+t \\\end{matrix}{ }(t\)是参数\()\),
\((\)Ⅰ\()\)写出直线\(l\)的普通方程和曲线\(C\)的直角坐标方程;
\((\)Ⅱ\()\)设曲线\(C\)经过伸缩变换\({ }\!\!\{\!\!{ }\begin{matrix} {x}{{{'}}}=2x \\ {y}{{{'}}}=y \\\end{matrix}{ }\)得到曲线\({C}{{{'}}}\),曲线\({C}{{{'}}}\)任一点为\(M\left( x,y \right)\),求点\(M\)直线\(l\)的距离的最大值.
\((\)二\()\)设函数\(f\left( x \right){=}45{|}x{-}a{|}\).
\((1)\)当\(a{=}2\)时,解不等式\(f\left( x \right){\geqslant }7\mathrm{{-}45{|}x{-}1{|}}\)
\((2)\)若\(f\left( x \right){\leqslant }1\)的解集为\(\left\lbrack 0{,}2 \right\rbrack\),\(\dfrac{1}{m}{+}\dfrac{1}{2n}{=}a\left( m{ > }0{,}n{ > }0 \right)\),求\(m{+}4n\)的最小值.