If a line is tangent to one point and normal at other point on the curve $x=4t^2+3,y=8t^3-1$, then slope of such a line is

$x=4t^2+3;y=8t^3-1$

$\frac{dx}{dt}=8t,\frac{dy}{dt}=24t^2$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=3t$

Let the tangent at $P(4t^2+3,8t^3-1)$ be normal at $Q(4t_1^2+3,8t_1^3-1)$

Equation of tangent at P

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

Equation (i) passes through Q

$8(t_1^3-t^3)=3t(4t_1^2-4t^2)$

$8(t_1^2+t^2+t{t}_1)=12t(t_1+t)$

$(t_1-t)(t+2t_1)=0$

$t=-2t_1$

${\left(\frac{{\displaystyle dy}}{{\displaystyle dx}}\right)}_t=\frac{1}{{\left(\frac{dy}{dx}\right)}_{t_1}}\Rightarrow 3t=\frac{-1}{3t_1}$

$\Rightarrow 9t{t}_1=-1$

$t{t}_1=\frac{-1}{9}$

$-2t_1^2=\frac{-1}{9}\Rightarrow t_1^2=\frac{1}{18}$

$t_1=\pm \frac{1}{3\sqrt{2}}$

$=\frac{-1}{3t_1}=\pm \sqrt{2}$