Why is this solution incorrect? Prove that on the axis of any parabola there is a certain point

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 ${\displaystyle \frac{1}{P{K}^2}+\frac{1}{Q{K}^2}}$ 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 ${\displaystyle y^2=4ax}$.
let the point K be (c,0)
Equation of line PQ using parametric coordinates: ${\displaystyle x=c+r\cos \theta }$ (Equation 1), ${\displaystyle y=r\sin \theta }$ (Equation 2)
From equation 1 and 2: ${\displaystyle {(x-c)}^2+y^2=r^2}$ (Equation 3)
Using Equation 3 and ${\displaystyle y^2=4ax}$, we get this quadratic in r: ${\displaystyle r^2-4ax-{(x-k)}^2=0}$ (Equation 4)
Roots of this quadratic ${\displaystyle \left(r_1\text{and}{r}_2\right)}$ would be the lengths of PK and QK
From Equation 4, ${\displaystyle r_1+r_2=0}$ and ${\displaystyle r_1{r}_2=-(4ax+{(x-k)}^2)}$
We know, ${\displaystyle \frac{1}{P{K}^2}+\frac{1}{Q{K}^2}=\frac{1}{r_1^2}+\frac{1}{r_2^2}=\frac{{(r_1+r_2)}^2-2r_1{r}_2}{{\left(r_1{r}_2\right)}^2}=\frac{-2}{r_1{r}_2}}$
As value of ${\displaystyle r_1{r}_2}$ is not constant thus ${\displaystyle \frac{1}{P{K}^2}+\frac{1}{Q{K}^2}}$ does not turn out to be constant. Hence, this solution is incorrect.
I've seen the correct solution but I wanted to know why this solution is incorrect?