The slope of the straight line which is both tangent and normal to the curve $4x^3=27y^2$ is

$x=3t^2,y=2t^3,\frac{dy}{dx}=t$

The tangent at $t,y-2t^3=t\left(x-3t^2\right)\Rightarrow tx-y=t^3$

The normal at $t_1,t_{1}y+x=2t_1^4+3t_1^2$

Now, (i) and (ii) are identical, $\frac{t}{1}=\frac{-1}{t_1}=\frac{t^3}{2t_1^4+3t_1^2}$

$\Rightarrow -t^3=2t_1^3+3t_1$ and $t_1=-\frac{1}{t}\text{. }$

Eliminating $t_1$ we get $t^6=2+3t^2\Rightarrow t^2=2\Rightarrow t=\pm \sqrt{2}$