Let $P$ be $\left(a t^2, 2 a t\right)$ and $Q$ be $(h, k)$.
Also given, $A=(-2,3)$
Now, since $Q$ divides $A P$ in the ratio $3: 2$ i.e.,
$$
\frac{A Q}{Q P}=\frac{3}{2}
$$


$$
\begin{array}{rlrl}
\Rightarrow & \quad 5 k & =6 a t+6 \\
\Rightarrow \quad t & =\frac{5 k-6}{6 a} \text { put in Eq. (i), we get } \\
5 h & =3 a\left(\frac{5 k-6}{6 a}\right)^2-4 \\
\Rightarrow \quad 5 h+4 & =\frac{(5 k-6)^2}{12 a} \Rightarrow(5 k-6)^2=60 a h+48 a \\
\Rightarrow \quad(5 k-6)^2 & =12 a(5 h+4)
\end{array}
$$
Hence, the locus of $Q(h, k)$ is

On comparing Eq. (iii) by $Y^2=4 A X$, we get,
$$
\begin{aligned}
\quad & Y & =5 y-6, X=5 x+4 \text { and } 4 A=12 a \\
& A & =3 a
\end{aligned}
$$
So, focus of $Y^2=4 A X$ is $(A, 0)$ i.e. $X=A$ and $Y=0$ Hence, focus of Eq. (iii) is
$$
\begin{aligned}
& & 5 x+4 & =3 a \text { and } 5 y-6=0 \\
\Rightarrow & & x & =\frac{3 a-4}{5} \text { and } y=\frac{6}{5} \\
& \text { i.e. } & (x, y) & =\left(\frac{3 a-4}{5}, \frac{6}{5}\right)=\text { Focus }
\end{aligned}
$$