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 $\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}$ 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 ${y}^{2}=4ax$.

let the point K be (c,0)

Equation of line PQ using parametric coordinates: $x=c+r\mathrm{cos}\theta$ (Equation 1), $y=r\mathrm{sin}\theta$ (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 $\left({r}_{1}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{r}_{2}\right)$ would be the lengths of PK and QK

From Equation 4, ${r}_{1}+{r}_{2}=0$ and ${r}_{1}{r}_{2}=-(4ax+{(x-k)}^{2})$

We know, $\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}=\frac{1}{{r}_{1}^{2}}+\frac{1}{{r}_{2}^{2}}=\frac{{({r}_{1}+{r}_{2})}^{2}-2{r}_{1}{r}_{2}}{{\left({r}_{1}{r}_{2}\right)}^{2}}=\frac{-2}{{r}_{1}{r}_{2}}$

As value of $r}_{1}{r}_{2$ is not constant thus $\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}$ 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?