<s>[INST] Why is this solution incorrect? 
Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ of the parabola be drawn through it, then
$\frac{1}{PK^2}
 + 
\frac{1}{QK^2}$
is the same for all positions of the chord.

If it would be valid for standard parabola than it would be valid for all parabolas. Thus, proving for $y^2=4ax$.
let the point K be (c,0)
Equation of line PQ using parametric coordinates:
$x=c+rcos\theta$ (Equation 1), $y=rsin\theta$ (Equation 2)
From equation 1 and 2: $(x-c)^2+y^2=r^2$ (Equation 3)
Using Equation 3 and $y^2=4ax$, we get this quadratic in r: $r^2- 4ax -(x-k)^2=0$ (Equation 4)
Roots of this quadratic $(r_1 \ and \ r_2)$ would be the lengths of $PK$ and $QK$
From Equation 4, $r_1 + r_2=0$ and $r_1r_2=-(4ax+(x-k)^2)$
We know, $\frac{1}{PK^2}+\frac{1}{QK^2}=\frac{1}{r_1^2}+\frac{1}{r_2^2}=\frac{(r_1 + r_2)^2-2r_1r_2}{(r_1r_2)^2}=\frac{-2}{r_1r_2}$
As value of $r_1r_2$ is not constant thus $\frac{1}{PK^2}+\frac{1}{QK^2}$ does not turn out to be constant. Hence, this solution is incorrect.
I've seen the correct solution but I wanted to know $\mathbf {why \ this \ solution \ is \ incorrect?}$
 [/INST] 
Roots of this quadratic ($r_1$ and $r_2$) would be the lengths of PK and QK.

This part is incorrect. Roots of this equation correspond to the same value of $x$. P and Q can have different $x$.
Update:
$x=c+r \cos \theta$, $y=r \sin \theta$
$$y^2=4ax \Rightarrow r^2 \sin^2\theta - 4ar \cos\theta - 4ac=0$$
Absolute values of roots of this equation would be the PK and QK.
$r_1+r_2=\frac{4a\cos\theta}{\sin^2\theta}$, $r_1r_2=-\frac{4ac}{\sin^2\theta}$
$$\frac{1}{PK^2}+\frac{1}{QK^2}=\frac{(r_1+r_2)^2-2r_1r_2}{(r_1 r_2)^2}=\frac{16a^2\cos^2\theta+8ac\sin^2\theta}{16a^2c^2}$$
This value does not depend on $\theta$ at $c=2a$. This is the answer.
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