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.

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.
