<s>[INST] Find the minimun of $MN+\frac{3}{5}MP$, $MN$ and $MP$ is two sides of a quadrilateral. In a quadrilateral $OPMN$ ，$\angle NOP=90^\circ$,$ON=1$,$OP=3$, and $M$ satisfy $\vec{MO}\cdot\vec{MP}=4$, find the minimum of
$MP+\frac{3}{5}MN$

I choose the vertex $O$ of $OPMN$ as the origin of the coordinate system, and ON as the $x$-axis, $OP$ as the $y$-axis. 

so $N(1,0)$ and $P(0,-3)$ and assume $M(x,y)$ then $\vec{MO}=(-x,-y)$ $\vec{MP}=(-x,-3-y)$. then $$x^2+3y+y^2=4\Rightarrow x^2+\left(y+\frac{3}{2}\right)^2=\left(\frac{5}{2}\right)^2$$
I know trail of point $M$ is a circle centered at $(0,-\frac{3}{2})$. 
but I have no idea to go to next. and I use $x=\frac{5}{2}\cos
\theta, y=\frac{5}{2}\sin\theta -\frac{3}{2}$ and we have $$MP+\frac{3}{5}MN =f(\theta)= \sqrt{\frac{17}{2}+\frac{15}{2}\sin
\theta} + \frac{3}{5}\sqrt{\frac{19}{2}-5\cos\theta-\frac{15}{2}\sin\theta}$$
I use Mathematica to find the numerical minimum of $f(\theta)$. It is about $$f_{\min}(\theta)\approx f(-1.45162)=3.45245$$
it seems when $M$ near the bottom of the circle, $MP+\frac{3}{5}MN$ take the minimum . Please help me .thanks very much. :)
 [/INST] from Maple:
$y = -\frac{969+85 \sqrt{273} }{596} = -3.982265922$
Minimum of $MP+\frac{3}{5}MN$ is $\frac{\sqrt{298}}{5} = 3.452535300$
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