The slope of  the normal to the curve $x=a{t}^4,y=a{t}^5$ at $t=1$ is

$x=a.t^4\Rightarrow \frac{dx}{dt}=4a.t^3$  

$y=a.t^5\Rightarrow \frac{dy}{dt}=5a.t^4$ $\Rightarrow \frac{dy}{dx}=\frac{5}{4}t$

$m={\left(\frac{dy}{dx}\right)}_{t=1}=\frac{5}{4}$  

Slope of normal$=\frac{-4}{5}$