A line L : y = m x + 3 meets y - a x is at E(0,3) and the arc of...

tangent at F yt = c + 4t²
a : x = 0 y = 4t (0,4t)
(4t²,8t) satisfies the line
8t = 4mt² + 3
4mt² - 8t + 3 = 0
\text{Area} = \dfrac{1}{2} \left|\begin{cases}0 & 3 & 1 \\ 0 & 4 t & 1 \\ 4 {t}^{2} & 8 t & 1\end{cases}\right|

$= \dfrac{1}{2} \left(4 {t}^{2} \left(3 - 4 t\right)\right)$
$= 2 {t}^{2} \left(3 - 4 t\right)$
$\text{A} = 2 \left[3 {t}^{2} - 4 {t}^{3}\right]$
$= \dfrac{\text{dA}}{\text{dt}} = 2 \left[6 t - 1 2 {t}^{2}\right]$
= 24 t(1 - 2t)

t = 1/2 maxima
$\text{G} \left(0 4 t\right) ⇒ \text{G} \left(0 2\right)$
y₁ = 2
\left({x}_{0} {y}_{0}\right) = \left(4 {t}^{2} 8 t\right) = \left(1 4\right)$<br />${y}_{0} = 4$<br />$\text{Area} = 2 \left(\dfrac{3}{4} - \dfrac{1}{2}\right) = 2 \left(\dfrac{3 - 2}{4}\right) = \dfrac{1}{2}