回復 1# shmilyho 的帖子
\(P\left( 0,p \right),Q\left( q,0 \right),R\left( 1,3 \right),M\left( \frac{q}{2},\frac{p}{2} \right)\)
直線PR之斜率為\(3-p\)，直線PQ之斜率為\(\frac{1}{p-3}\)
\(\begin{align}
& \frac{1}{p-3}=\frac{p-0}{0-q} \\
& q=3p-{{p}^{2}} \\
& M\left( \frac{3p-{{p}^{2}}}{2},\frac{p}{2} \right) \\
& \frac{3p-{{p}^{2}}}{2}+\frac{4{{\left( \frac{p}{2}-\frac{3}{4} \right)}^{2}}-\frac{9}{4}}{2}=0 \\
& x+\frac{4{{\left( y-\frac{3}{4} \right)}^{2}}-\frac{9}{4}}{2}=0 \\
& 2{{y}^{2}}-3y+x=0 \\
\end{align}\)
