Normal of parabola $y^2=4ax$

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A normal is drawn to a parabola $y^2=4ax$ at any point other than the vertex and it cuts the parabola again at a point whose distance from the vertex is not less than $\lambda\sqrt{6}a$, then the value of $\lambda$ is?

My attempt,

Let $t_1$ be point on parabola on which normal is drawn and it intersects again it at $t_2$, then its distance from origin will be $$at_2\sqrt{t_2^2+4}$$

Given, $at_2\sqrt{t_2^2+4}\geq\lambda\sqrt{6}a$ and $t_2=-t_1-\dfrac{2}{t_1}$.

So how do I minimize $t_1$ parameter for these lengths? please help.

Or any other elegant method is appreciated.