The curve $C$ has parametric equations
$x = 2t - \frac{1}{t^2}, \quad y = 2t + \frac{1}{t}, \quad t \in \mathbb{R}, t \ne 0.$
The point $P$ on the curve has parameter $t = 1$.

1. Find the coordinates of point $P$ by substituting $t=1$.

$x = 2(1) - \frac{1}{1^2} = 1$
$y = 2(1) + \frac{1}{1} = 3$
So, $P$ is $(1, 3)$.

2. Find the derivatives of $x$ and $y$ with respect to $t$.

$x = 2t - t^{-2} \implies \frac{dx}{dt} = 2 + 2t^{-3} = 2 + \frac{2}{t^3}$
$y = 2t + t^{-1} \implies \frac{dy}{dt} = 2 - t^{-2} = 2 - \frac{1}{t^2}$

3. Find the gradient of the tangent, $\frac{dy}{dx}$, at $t = 1$.

At $t = 1$:
$\frac{dx}{dt} = 2 + 2 = 4$
$\frac{dy}{dt} = 2 - 1 = 1$

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1}{4}$

4. Find the equation of the tangent at $P$.

Using the point $(1, 3)$ and gradient $m = \frac{1}{4}$:
$y - 3 = \frac{1}{4}(x - 1)$
$4y - 12 = x - 1$
$4y - x = 11$

5. Find the equation of the normal at $P$.

The gradient of the normal is $-\frac{1}{m} = -4$.
$y - 3 = -4(x - 1)$
$y - 3 = -4x + 4$
$y + 4x = 7$

1. Retrieve the coordinates of $P$ and the equations of the tangent and normal from part (i)

(Self-contained derivation for completeness)
At $t = 1$, $x = 2(1) - \frac{1}{1^2} = 1$ and $y = 2(1) + \frac{1}{1} = 3$, giving $P(1, 3)$.
$\frac{dx}{dt} = 2 + \frac{2}{t^3} = 4$ and $\frac{dy}{dt} = 2 - \frac{1}{t^2} = 1$ at $t = 1$, so the gradient of the tangent is $\frac{dy}{dx} = \frac{1}{4}$.

Equation of tangent at $P$:
$y - 3 = \frac{1}{4}(x - 1) \implies y = \frac{1}{4}x + \frac{11}{4}$

Equation of normal at $P$:
$y - 3 = -4(x - 1) \implies y = -4x + 7$

2. Find the coordinates of the points $A$ and $B$

The tangent meets the $y$-axis at $B$. Substituting $x = 0$ into the tangent equation:
$y = \frac{11}{4} \implies B\left(0, \frac{11}{4}\right)$

The normal meets the $x$-axis at $A$. Substituting $y = 0$ into the normal equation:
$0 = -4x + 7 \implies x = \frac{7}{4} \implies A\left(\frac{7}{4}, 0\right)$

3. Calculate the area of the quadrilateral $OAPB$

We can split quadrilateral $OAPB$ into two triangles: $\triangle OAP$ and $\triangle OPB$.

Area of $\triangle OAP = \frac{1}{2} \times \text{base } OA \times \text{perpendicular height of } P$ from the $x$-axis
$= \frac{1}{2} \left(\frac{7}{4}\right) (3) = \frac{21}{8}$

Area of $\triangle OPB = \frac{1}{2} \times \text{base } OB \times \text{perpendicular height of } P$ from the $y$-axis
$= \frac{1}{2} \left(\frac{11}{4}\right) (1) = \frac{11}{8}$

Total Area $= \frac{21}{8} + \frac{11}{8} = \frac{32}{8} = 4$