Parabolas within parabolas

The answer is $\frac{4}{3^{10}}$ . here is the solution.

Since tangents are at right angles, the line joining the point of contacts is the focal chord.

here $a=1$

Let $A$ $(t_{1}^{2},2t_{1})$ and $B$ $(t_{2}^{2},2t_{2})$ .

Since these are end points of focal chord $t_{1}t_{2}=-1$

let $t_{1}=t$

Point $A(t^{2},2t)$ and $B(\frac{1}{t^{2}},-\frac{2}{t})$

Let centroid $C(h,k)$ and given $K(0,0)$

$\huge{h=\frac{t^{2}+\frac{1}{t^{2}}+0}{3}}$ ....... $(1)$

and

$\huge{k=\frac{2t-\frac{2}{t}+0}{3}}$

=> $\large{t-\frac{1}{t}=\frac{3k}{2}}$

From $(1)$

$3h-2=(t-\frac{1}{t})^{2}$

$\large{P_{1}= y^{2}=\frac{4}{3}(x-\frac{2}{3})}$

on repeating similar procedure

$\large{P_{2}=y^{2}=\frac{4}{3^{2}}(x-\frac{4}{9})}$ thus

$\large{P_{10}=y^{2}=\frac{4}{3^{10}}(x-\frac{2^{10}}{3^{10}})}$

$\large{\boxed{a+b=4+3^{10}=59053}}$