The complete set of value of 'a' for which there exists at least one line that is tangent to the graph of the curve $y=x^3-a x$ at one point and normal to the graph at another point is given by

$[4 / 3, \infty)$

The coordinates of any point P on $y=x^3-a x$ are $\left(t, t^3-a t\right)$ and the equation of the tangent at P is $y=\left(3 t^2-a\right) x-2 t^3$.

The abscissae of the points of intersection of $y=x^3-a x$ and the tangent $y=\left(3 t^2-a\right) x-2 t^3$ are the roots of the equation

$\left(3 t^2-a\right) x-2 t^3 =x^3-a x$

$\Rightarrow x^3-3 t^2 x+2 t^3 =0 \Rightarrow(x-t)^2(x+2 t)=0 \Rightarrow x=-2 t, t$

Clearly, x = t corresponds to point P. So, the tangent at P cuts the curve again at $Q\left(-2 t,-8 t^3+2 a t\right)$.

Now,

$y=x^3-a x \Rightarrow \frac{d y}{d x}=3 x^2-a \Rightarrow\left(\frac{d y}{d x}\right)_Q=12 t^2-a$

So, the slope of the normal at Q is $-\frac{1}{\left(\frac{d y}{d x}\right)_Q}=-\frac{1}{12 t^2-a}$

It is given that the tangent at P is normal at Q.

∴  $\left(\frac{d y}{d x}\right)_P=-\frac{1}{\left(\frac{d y}{d x}\right)_Q} \Rightarrow\left(3 t^2-a\right)=-\frac{1}{12 t^2-a}$

$\Rightarrow \left(3 t^2-a\right)\left(12 t^2-a\right)+1=0$

$\Rightarrow 36 t^4-15 a t^2+\left(a^2+1\right)=0$

$\Rightarrow 36 u^2-15 a u+\left(a^2+1\right)=0$,  where  $t^2=u$

This equation must have at least one real root u and $t^2=u$ implies that the root must be positive.

∴  $\frac{15 a}{72}>0$ and $225 a^2-144\left(a^2+1\right) \geq 0$         [Using : $-\frac{b}{2 a}>0$ and Disc c ≥ 0]

$\Rightarrow a>0$  and  $9(3 a-4)(3 a+4) \geq 0$

$\Rightarrow a \geq \frac{4}{3} \Rightarrow a \in[4 / 3, \infty)$