The area of triangle formed by tangent and normal at (8, 8) on the parabola ${\mathrm{y}}^2=8\mathrm{x}$ and the axis of the parabola is

${\mathrm{y}}^2=8\mathrm{x}$

$\mathrm{a}=2,$ 

Given point =(8, 8)

diff w.r.to x on both sides 

$$\begin{gathered} 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=8 \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{\mathrm{y}} \end{gathered}$$

slope of tangent at (8, 8) is $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ 

Eq of tangent is

$$\begin{gathered} \mathrm{y}-8=\frac{1}{2}(\mathrm{x}-8) \\ 2\mathrm{y}-16=\mathrm{x}-8 \\ \mathrm{x}-2\mathrm{y}+8=0 -① \end{gathered}$$

Eq of normal is

$$\begin{gathered} \mathrm{y}-8=-2(\mathrm{x}-8) \\ \mathrm{y}-8=-2\mathrm{x}+16 \\ 2\mathrm{x}+\mathrm{y}-24=0 -② \end{gathered}$$

Eq of axe is $\mathrm{y}=0$ vertices of triangle are $(-8, 0) (12, 0) (8, 8)$

$$\begin{gathered} ∆=\frac{1}{2}\left|{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right| \\ ∆=\frac{1}{2}\left|-8\left(-8\right)+12\left(8\right)+8\left(0\right)\right| \\ ∆=\frac{1}{2}\left|64+96\right|=80 \mathrm{sq} \end{gathered}$$