If the three normals drawn to the parabola, y ^ 2 = 2x pass through...

Let the equation of the normal be $y = mx - 2am - a{m}^{3}$

Here $4a = 2$

$a = \dfrac{1}{2}$

$y = mx - m - \left(\dfrac{1}{2}\right){m}^{3}$

It passes through $A ( a, 0 )$ then

$0 = am - m - \left(\dfrac{1}{2}\right){m}^{3}$

$m = 0, {m}^{2} - 2 ( a - 1 ) = 0$

For real values of $m$

$2 ( a - 1 ) > 0$

$⇒ a > 1$