Step 1:
Tangent to the parabola $y^2=8x$ at $t=\large\frac{1}{2}$
Here $4a=8$
$\Rightarrow a=2$
Step 2:
Tangent at $'t'$ is given by $yt=x+at^2$
$yt=x+2t^2$
For $t=\large\frac{1}{2}$
$y\times \large\frac{1}{2}=$$x+2\times \large\frac{1}{4}$
(i.e) $\large\frac{y}{2}=$$x+\large\frac{1}{2}$
On simplifying we get
$2x-y+1=0$
Step 3:
The normal at $'t'$ is given by
$tx+y=2at+at^3$
$tx+y=4t+2t^3$
For $t=\large\frac{1}{2}$
$\large\frac{x}{2}$$+y=\large\frac{4}{2}+$$2\times \large\frac{1}{8}$
$\large\frac{x}{2}$$+y=\large\frac{9}{4}$
On simplifying we get
$2x+4y-9=0$
Step 4:
Equation of tangent : $2x-y+1=0$
Equation of normal : $2x+4y-9=0$