Equation of the normal to a parabola $${y^2} = 4bx$$ at point

$$\left( {bt_1^2,2b{t_1}} \right)$$ is $$y = - {t_1}x + 2b{t_1} + bt_1^3$$

As given, it also passes through $$\left( {bt_2^2,2b{t_2}} \right)$$ then

$$2b{t_2} = {t_1}bt_2^2 + 2b{t_1} + bt_1^3$$

$$2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right)$$

$$ = - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right)$$

$$ \Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right)$$

$$ \Rightarrow {t_2} + {t_1} = - {2 \over {{t_1}}}$$

$$ \Rightarrow {t_2} = - {t_1} - {2 \over {{t_1}}}$$

Any tangent to the parabola $${y^2} = 8ax$$ is

$$y = mx + {{2a} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

If $$(i)$$ is a tangent to the circle, $${x^2} + {y^2} = 2{a^2}$$ then,

$$\sqrt {2a} = \pm {{2a} \over {m\sqrt {{m^2} + 1} }}$$

$$ \Rightarrow {m^2}\left( {1 + {m^2}} \right) = 2$$

$$ \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0$$

$$ \Rightarrow m = \pm 1.$$

So from $$(i),$$ $$y = \pm \left( {x + 2a} \right).$$