2023-I-16

$\left\{ \begin{array}{ll}
p+5p=\dfrac{-a}{1} & \ldots \unicode{x2460} \\
p\times 5p =\dfrac{b}{1} & \unicode{x2461}
\end{array}\right.$

From $\unicode{x2460}$, we have

$\begin{array}{rcl}
6p & = & -a \\
p & = & \dfrac{-a}{6}
\end{array}$

Sub. $p=\dfrac{-a}{6}$ into $\unicode{x2461}$, we have

$\begin{array}{rcl}
5\left(\dfrac{-a}{6}\right)^2 & = & b \\
5a^2 & = & 36b
\end{array}$

Sub. $\unicode{x2461}$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
x^2 +(mx)^2 -6x-12(mx)+20 & = & 0 \\
(1+m^2)x^2 +(-6-12m)x +20 & = & 0 \\
x^2 +\dfrac{-6-12m}{1+m^2}x +\dfrac{20}{1+m^2} & = & 0 \ \ldots (*)
\end{array}$

Let $(q_1,q_2)$ and $(r_1,r_2)$ be the coordinates of $Q$ and $R$ respectively.

Since $OQ:QR =1:4$, then we have

$\begin{array}{rcl}
\dfrac{r_1}{1+4} & = & q_1 \\
r_1 & = & 5q_1
\end{array}$

Hence, the roots of $(*)$ are $q_1$ and $5q_1$.

Then by the result of (a), we have

$\begin{array}{rcl}
5\left(\dfrac{-6-12m}{1+m^2}\right)^2 & = & 36 \left(\dfrac{20}{1+m^2}\right) \\
5[-6(1+2m)]^2 & = & 720(1+m^2) \\
(1+2m)^2 & = & 4(1+m^2) \\
4m^2+4m+1 & = & 4+4m^2 \\
4m & = & 3 \\
m & = & \dfrac{3}{4}
\end{array}$