Minimum slope of a chord to parabola A line is drawn from  $(-2,0)$ to intersect
$y^2 = 4x \,\,$ in P,  Q within the first quadrant, such that 
$$ \frac{1}{AP} +\frac{1}{AQ} < \frac{1}{4} $$
Find the minimum value of the line slope. A is the origin.
I had basically let the coordinates of the parabola in parametric form and then used the given condition, but couldn't find the minimal slope.

As it is stated  in the  current form, this problem - as shown below - has no real solutions. I suspect that the original formulation might have been slightly different, in particular with $A$ indicating the point $(-2,0)\,\,$ and maybe with a more general equation of the parabola (again see below). In this answer, I will assume that $A=(-2,0)\,\,$ and that $P$ and $Q$ are distinct points. Also, I will firstly consider the case of the specific parabola described in the OP, and then the case of a more general parabola equation. Lastly, I will add a brief comment on the case where $A$ is the origin.
Since the line crosses the point $(-2,0)\,\,$, it has the form 
$$y=ax+2a$$
The intersection points of this line with the parabola have $x$-coordinates given by
$$ \frac{2 (1 - a^2 \pm \sqrt{1 - 2 a^2})}{a^2}$$
whereas the $y$-coordinates are
$$ \frac{2 (1 - a^2 \pm \sqrt{1 - 2 a^2})}{a}+2a$$
Calculating the length of $\overline{AP}$ using standard formulas yields 
$$\overline{AP}= \left[\left( \frac{2 (1 - a^2 - \sqrt{1 - 2 a^2})}{a^2} \right)^2+ \left(\frac{2 (1 - a^2 - \sqrt{1 - 2 a^2})}{a}+2a \right)^2 \right]^{1/2}$$
which can be simplified as 
$$\overline{  AP}=\frac{2 \sqrt{2 - 2 a^2 - a^4 -2 \sqrt{1 - 2 a^2}}}{a^2}$$
In a similar manner we get
$$\overline { AQ}=\frac{2 \sqrt{2 -2 a^2- a^4 + 2 \sqrt{1 - 2 a^2}}}{a^2}$$
Note that since we must have $1 - 2 a^2 >0\,\,\,$ to have two separate intersection points, we get the condition
$$-\frac{1}{\sqrt{2}} < a <   \frac{1}{\sqrt{2}}  $$
to obtain a different real value for $\overline{AP}$ and $\overline {AQ}$. 
Substituting the expressions above in 
$$\frac{1}{\overline{AP}}+\frac{1}{\overline{AQ}}<\frac{1}{4}$$
and simplifying, we get
$$\frac{1}{2 \sqrt{1 + a^2}}<\frac{1}{4}$$
whose positive solution is
$$a>\sqrt{3}$$
Because this solution has no intersection with the range of $a$ needed to have $AP$ and $AQ$ real,  we conclude that there are no real values of $a$ that satisfy the conditions stated in the OP.
Notably, after some searching, I found on Google a nearly identical version of this problem with no solution but four possible answers ($<\sqrt{3}\,\,$, $>\sqrt{3}\,\,$,   $\geq \sqrt{3}\,\,$, none of them),   with the only difference that the initial parabola has equation $y^2=4kx \,\,$. The point $A$ is explicitly defined as the point $(-2,0)\,\,$. This more general version of the problem is interesting because, after calculations that are similar to those shown above, it leads to the same final inequality $1/(2 \sqrt{1 + a^2})<1/4\;$ with solution $a>\sqrt{3}\,\,\,$ (in other words, in the final part of the calculations and simplifications, $k$ is canceled out). In contrast, $k$ is not canceled out in the condition necessary for $AP$ and $AQ$ to be distinct and real, which becomes 
$$-\sqrt{\frac{k}{2}} < a <   \sqrt{\frac{k}{2}}  $$
allowing to find, for $k>6\,$, an intersection between this range and that stated in the solution, and giving the answer $a>\sqrt{3}\,$ as the inferior bound of the slope. Note that such condition, for $k=1\,\,$, is the same obtained above considering the problem for the specific parabola $y^2=4x\,\,$.
Lastly, repeating the same calculations for the case where $A$ is the origin leads to similar results. The final inequality in this case yields the numerical solution $a>1.23414...\;$ (apparently with no closed form). Again, because of the condition needed for $\overline{AP}$ and  $\overline{AQ}$  to be real and distinct, there are no real solutions for the case with parabola equation $y^2=4x\,\,$. On the other hand, in the general case with parabola equation $y^2=4kx\,\,$, there are real solutions only for 
$$k>2\cdot 1.23414...^2=3.04620...\,\,\,\,$$ 
and the numerical solution $a>1.23414...\,\,\,$ is the lower bound of the slope.
