略

设点 $M$ 的坐标是 $\left( {x,y} \right)$，$P$ 的坐标是 $\left({x_p},{y_p}\right)$，
则根据题意得\[{x_p} = x,{y_p} = \dfrac{5}{4}y,\]因为点 $P $ 在圆 ${x^2} + {y^2} = 25$ 上，所以\[{x^2} + {\left( {\dfrac{5}{4}y} \right)^2} = 25,\]整理即得轨迹 $C $ 的方程为\[\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{{16}} = 1.\]

略

过点 $\left(3.0\right)$ 且斜率为 $\dfrac{4}{5}$ 的直线方程是\[y = \dfrac{4}{5}\left(x - 3\right),\]设此直线与 $C$ 的交点为 $A\left({x_1},{y_1}\right)$，$B\left({x_2},{y_2}\right)$，将 $y = \dfrac{4}{5}\left(x - 3\right)$ 代入 轨迹 $C$ 的方程得\[\dfrac{{{x^2}}}{{25}} + \dfrac{{{{\left(x - 3\right)}^2}}}{{25}} = 1,\]化简得\[{x^2} - 3x - 8 = 0,\]解得\[\begin{split}{x_1} &= \dfrac{{3 - \sqrt {41} }}{2} , \\ {x_2} &= \dfrac{{3 + \sqrt {41} }}{2},\end{split}\]所以线段 $AB$ 的长度是\[\begin{split}\left| {AB} \right|&{\text{ = }}\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} \\& = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + \dfrac{{16}}{{25}}{{\left( {{x_1} - {x_2}} \right)}^2}}\\& = \sqrt {\left( {1 + \frac{{16}}{{25}}} \right){{\left( {{x_1} - {x_2}} \right)}^2}}\\& = \sqrt {\frac{{41}}{{25}} \times 41} = \frac{{41}}{5}.\end{split}\]故所截线段的长度是 $\dfrac{{41}}{5}$．