Find the equation of the tangent and normal to the parabola $y^{2}=8x$ at $t=\large\frac{1}{2}$

Toolbox:
• The equation of tangent at $'t'$ is given by $yt=x+at^2$
• The equation of normal at $'t'$ is given by $y+tx=2at+at^3$
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}$