Find tangent and normal equations with a scale parameter

$dx/dt=-a/t^2$, $dy/dt=a(1+2/t^3)$. For $a\ne0$, $dy/dx=-t^2-2/t$. The printed question does not explicitly exclude $a=0$; then the curve collapses to the origin and has no uniquely defined tangent or normal. The following requested lines use the intended nondegenerate case $a\ne0$.

The point is $(3a/2,7a/4)$ and tangent gradient is $-5$, so the normal gradient is $1/5$. Thus $y-7a/4=(x-3a/2)/5$, simplifying to $y=x/5+29a/20$.

For $x=0$ and $a\ne0$, $1+1/t=0$, hence $t=-1$, which gives $y=-2a$. The gradient is $-(-1)^2-2/(-1)=1$. Thus $y+2a=x$, or $y=x-2a$.