jmc

Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix. Completing the square on $x,$ we get $$y=\frac{5}{4}{\left(x-\frac{2}{5}\right)}^2+\frac{3}{10}.$$To make the algebra a bit easier, we can find the directrix of the parabola $y=\frac{5}{4}{x}^2,$ shift the parabola right by $\frac{2}{5}$ units to get $y=\frac{5}{4}{\left(x-\frac{2}{5}\right)}^2,$ and then shift it upward $\frac{3}{10}$ units to find the focus of the parabola $y=\frac{5}{4}{\left(x-\frac{2}{5}\right)}^2+\frac{3}{10}.$

Since the parabola $y=\frac{5}{4}{x}^2$ is symmetric about the $y$-axis, the focus is at a point of the form $(0,f).$ Let $y=d$ be the equation of the directrix.



Let $\left(x,\frac{5}{4}{x}^2\right)$ be a point on the parabola $y=\frac{5}{4}{x}^2.$ Then $$P{F}^2=x^2+{\left(\frac{5}{4}{x}^2-f\right)}^2$$and $P{Q}^2={\left(\frac{5}{4}{x}^2-d\right)}^2.$ Thus, $$x^2+{\left(\frac{5}{4}{x}^2-f\right)}^2={\left(\frac{5}{4}{x}^2-d\right)}^2.$$Expanding, we get $$x^2+\frac{25}{16}{x}^4-\frac{5f}{2}{x}^2+f^2=\frac{25}{16}{x}^4-\frac{5d}{2}{x}^2+d^2.$$Matching coefficients, we get $$\begin{array}{rl}1-\frac{5f}{2}& =-\frac{5d}{2},\\ f^2& =d^2.\end{array}$$From the first equation, $f-d=\frac{2}{5}.$ Since $f^2=d^2,$ $f=d$ or $f=-d.$ We cannot have $f=d,$ so $f=-d.$ Then $2f=\frac{2}{5},$ so $f=\frac{1}{5}.$

Then the focus of $y=\frac{5}{4}{x}^2$ is $\left(0,\frac{1}{5}\right),$ the focus of $y=\frac{5}{4}{\left(x-\frac{2}{5}\right)}^2$ is $\left(\frac{2}{5},\frac{1}{5}\right),$ and the focus of $y=\frac{5}{4}{\left(x-\frac{2}{5}\right)}^2+\frac{3}{10}$ is $\boxed{\left(\frac{2}{5},\frac{1}{2}\right)}.$