Given equation of parabolas
$$
y^2=5 x
$$
and $\quad x^2=5 y$
Now, $\quad x^2=5 y$
$\Rightarrow \quad y=\frac{x^2}{5}$
On substituting $y=\frac{x^2}{5}$ in Eq. (i), we get
$$
\left(\frac{x^2}{5}\right)^2=5 x
$$

$$
\Rightarrow \quad \frac{x^4}{25}=5 x
$$

$$
\begin{aligned}
& \Rightarrow & x^4 & =125 x \\
& \Rightarrow & x^4-125 x & =0 \\
& \Rightarrow & x\left(x^3-125\right) & =0
\end{aligned}
$$
On solving Eqs. (i) and (ii), we get
$$
\begin{array}{ll}
x=0 & \text { or } x=5 \\
y=0 & \text { or } y=5
\end{array}
$$
Hence, the point of intersection $(0,0)$ and $(5,5)$ lie on, line $x-y=0$.