If the normal substends a right angle at the focus of the parabola $y^2=8x$ then its length is taken in integer approximate value = ______

$S(2,0)$

Slope of $SP\times SQ=1$

$$\begin{gathered} \frac{2a{t}_1}{a{t}_1^2-2}\times \frac{2a{t}_2}{a_2^2-2}=-1 \\ \boxed{a=2} \\ \frac{4t_1}{2(t_1^2-1)}\times \frac{4t_2}{2(t_2^2-1)}=-1 \\ 4t_1{t}_2=-\left((t_1^2-1)(t_2^2-1)\right) \\ {t}_1^2{t}_2^2+2t_1{t}_2+1=t_1^2+t_2^2-2t_1{t}_2 \\ {t}_1{t}_2+t_2=t_1-1 \\ {t}_2=\frac{t_1-1}{t_1+1} \\ \boxed{t_2=-t_1-\frac{2}{t_1}} \end{gathered}$$

Solve

$$\begin{gathered} t_1=2 \\ {t}_2=3 \\ P(8,-8),Q(18,12) \\ P{Q}^2={10}^2+{20}^2=500 \\ PQ=10\sqrt{5} \end{gathered}$$