Let the tangent to the parabola S: y ^ 2 = 2x at the point P(2, 2)...

We have,

${y}^{2} = 2x$

Tangent at $$P\left(2, 2\right)$ is

$y\left(2\right) = 2\left(\dfrac{1}{2}\right)\left(x + 2\right)$

$⇒ 2y = x + 2$

It meets $x -$axis at $y = 0$ i.e., $x = - 2$

$∴ Q ≡ \left(- 2, 0\right)$

Normal at $P$ is

$⇒ y - 2 = - 2\left(x - 2\right)$

$⇒ y = 6 - 2x$

$∴$ Solving with ${y}^{2} = 2x$, we get

${\left(6 - 2x\right)}^{2} = 2x$

$⇒ 4{x}^{2} + 36 - 24x = 2x$

$⇒ 4{x}^{2} - 26x + 36 = 0$

$⇒ 2{x}^{2} - 13x + 18 = 0$

$⇒ x = \dfrac{13 \pm \sqrt{169 - 144}}{4} = \dfrac{13 \pm 5}{4}$

$⇒ x = \dfrac{9}{2}, \dfrac{1}{2}$

$∴ R ≡ \left(\dfrac{9}{2}, - 3\right)$

∴ Ar\left(\delta PQR\right) = \dfrac{1}{2}\left|\begin{cases}2 & 2 & 1 \\ - 2 & 0 & 1 \\ \dfrac{9}{2} & - 3 & 1\end{cases}\right|

$⇒ Ar\left(\delta PQR\right) = \dfrac{1}{2}\left[2 \cdot 3 - 2\left(- 2 - \dfrac{9}{2}\right) + 6\right]$

$⇒ Ar\left(\delta PQR\right) = \dfrac{1}{2}\left[12 + 13\right]$
$⇒ Ar\left(\delta PQR\right) = \dfrac{25}{2} sq.units