Since, the axis of the parabola is parallel to y-axis. Then, $(x-h)^2=4 a(y-k)$
Since, $e q^{\mathrm{n}}$ (i) passes through the given points $\left(0, \frac{2}{5}\right)$, $(4,-2) \&\left(1, \frac{8}{5}\right)$
$$
\therefore \mathrm{h}^2=4 \mathrm{a}\left(\frac{2}{5}-\mathrm{k}\right)
$$
$(4-h)^2=4 a(-2-k)$
$$
(1-h)^2=4 a\left(\frac{8}{5}-k\right)
$$
Solving eqn. (ii), (iii) \& (iv), we get
$$
\mathrm{h}=\frac{3}{2}, \mathrm{k}=\frac{7}{4}, \mathrm{a}=-\frac{5}{12}
$$
Putting above values in eqn. (i), we get
$$
\left(x-\frac{3}{2}\right)^2=-\frac{5}{3}\left(y-\frac{7}{4}\right)
$$
Let's check the point $\left(2, \frac{8}{5}\right)$ on equation (v):
$\because \quad \frac{1}{4}=\frac{1}{4} ; \therefore\left(2, \frac{8}{5}\right)$ lies on the parabola.