The coordinates of a point on the curve $x=4 t^2+3, y=8 t^3-1$ such that the tangent to the curve at that point is normal to the curve at its other intersection point with the curve, are

$\left(\frac{35}{9}, \pm \frac{16 \sqrt{2}}{27}-1\right)$

We have,

$x=4 t^2+3$ and $y=8 t^3-1 \Rightarrow \frac{d y}{d x}=3 t$

The equation of the tangent at any point 't' is

$y-8 t^3+1=3 t\left(x-4 t^2-3\right)$

Suppose it meets the curve again at the point '$t_1$'. Then,

$8 t_1{ }^3-1-8 t^3+1=3 t\left(4 t_1{ }^2+3-4 t^2-3\right)$

$\Rightarrow 8\left(t_1{ }^3-t^3\right)=12 t\left(t_1{ }^2-t^2\right)$

$\Rightarrow 4\left(t_1-t\right)\left\{2 t_1{ }^2+2 t t_1+2 t^2-3 t t_1-3 t^2\right\}=0$

$\Rightarrow 4\left(t_1-t\right)\left(2 t_1{ }^2-t t_1-t^2\right)=0$

$\Rightarrow \quad 4\left(t_1-t\right)^2\left(2 t_1+t\right)=0 \Rightarrow t_1=-\frac{t}{2}$

The slope of the normal at $t_1$ is $-\frac{1}{3 t_1}$

∴  $3 t=-\frac{1}{3 t_1}$

$\Rightarrow 9 t t_1=-1$

$\Rightarrow -9 t^2=-2$                  [∵ t₁ = -t/2]

$\Rightarrow t= \pm \frac{\sqrt{2}}{3}$

Hence, the required points are $\left(\frac{35}{9}, \pm \frac{16 \sqrt{2}}{27}-1\right)$