A variable parabola touches the $x$-axis and $y$-axis at $A(1,0)$ and $B(0,1)$. Find the locus of its focus. 
A variable parabola touches the $x$-axis and $y$-axis at $A(1,0)$ and $B(0,1)$ on the co-ordinate plane respectively. Now, we are required to find the locus of the focus of this variable parabola.

The process to arrive at this locus is a standard one, and goes like,
Starting with facts (observations),


*

*The parabola has $x$- and $y$-axes as its tangents, and it lies in the first quadrant

*We know that these tangents intersect orthogonally and hence the intersection point lies on its directrix.

*Since the directrix passes through the origin let its equation be $y=mx$.

*Now $A(1,0)$ and $B(0,1)$ lies on the parabola hence if we define focus as $F(h,k)$ we find that from the definition of parabola 
$$\begin{align}
FA &= \text{(distance from $A$ to the directrix)} \\
FB &=\text{(distance from $B$ to the directrix)}
\end{align}$$
Hence we have sufficient conditions to get the locus,


Writing, $$ (FA)^2 = (h-1)^2 + (k-0)^2 = \frac{|(0)-m(1)|^2}{1+m^2}$$
$$ (FB)^2 = (h-0)^2 + (k-1)^2 = \frac{|(1)-m(0)|^2}{1+m^2}$$
Adding both and simplifying we get the locus of $F(h,k)$ as, 
$$x^2 + y^2 - x - y + 0.5 = 0$$ 
This is an imaginary equation which doesn't give the locus of $F(h,k)$, So my question is how to interpret this result, What does it mean to have a set of imaginary focal points? or Are there is any reason to claim that my solution process is wrong? If yes, then what is the correct way to obtain the locus of $F(h,k)$?

Contrary to what you wrote, $x=y=\frac12$ satisfies your last equation: the circle isn’t imaginary but consists of a single point. This is as it should be: two points and the tangents at those points uniquely determine a parabola.  
I’ve seen a version of this problem in which the parabola is tangent to the coordinate axes, but at a pair of unspecified points. Perhaps that’s what was intended here.
