Ans: C
Let $a=\dfrac{kb}{c^2}$, where $k$ is a non-zero constant.
Let $a=\dfrac{kb}{c^2}$, where $k$ is a non-zero constant.
For $b=6$, $c=3$ and $a=-2$, we have
$\begin{array}{rcl}
-2 & = & \dfrac{k(6)}{(3)^2} \\
k & = & -3
\end{array}$
Therefore, $a = \dfrac{-3b}{c^2}$.
For $a=-9$ and $c=4$, we have
$\begin{array}{rcl}
-9 & = & \dfrac{-3b}{4^2} \\
b & = & 48
\end{array}$
