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.
 A: 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.
A: Convenient to shift reference to origin
$$y^2= 4(x-2) $$
Convenient to use polar coordinates
$$(x,y)= r ( \cos \theta, \sin \theta) $$
Plug in
$$ r^2 \sin^2 \theta =4 ( r \cos \theta-2)$$ 
Simplify, let me abbreviate for easier Latex typing 
$$ 4 r s_t^2 -4 r c_t +8 =0 $$
If roots are $r_1,r_2 (AP,AQ) $ for quadratic, sum and product are
$$ r_1+r_2= 4 c_t/s_t^2$$
and
$$ r_1\cdot r_2= 8/ s_t^2$$
Trick is to divide getting
$$ \frac{1}{ r_1} +  \frac{1}{ r_2} = c_t/2 $$
( BTW harmonic mean of the two segments $ AP,AQ = 4 \sec \theta $ )
Now if $$ c_t/2<\frac14$$
then for maximum slope
$$ t= \pi/3, m= slope= \sqrt3$$ 
Next find slope of tangent onto parabola from origin for maximum possible slope, we find intersections between
$$ y^2= 4(x-2) ,\quad y=mx$$
Plug in and get quadratic in, say $x$.
$$ m^2x^2 -4x +8=0$$
Discriminant is zero for tangent condition
$$ \Delta= 16- 4\cdot m^2\cdot 8 =0 $$
of tangent line L2
$$ m= tangent slope = \pm \frac{1}{\sqrt{2}}$$
For given slope line L1
$$ \sqrt3 > \frac {1}{\sqrt 2}$$
So given inequality can never be satisfied with the real numbers.
All the above are included together in a single graph below:

For line $L1$ neither the segments nor their harmonic mean exist in reals.
