The parametric equations of a curve are $x = t + 2$, $y = \dfrac{t}{2t - 1}$, where $t \in \mathbb{R}$, $t \ne \dfrac{1}{2}$.

1. Convert the parametric equations to a Cartesian equation.

From $x = t + 2$, we can express $t$ in terms of $x$:
$t = x - 2$

Substitute this into the equation for $y$:
$y = \frac{t}{2t - 1} = \frac{x - 2}{2(x - 2) - 1}$
$y = \frac{x - 2}{2x - 5}$

2. Find the coordinates of the points where the curve crosses the axes.

For the $x$-axis intersection, set $y = 0$:
$\frac{x - 2}{2x - 5} = 0 \implies x - 2 = 0 \implies x = 2$
The curve crosses the $x$-axis at $(2, 0)$.

For the $y$-axis intersection, set $x = 0$:
$y = \frac{0 - 2}{2(0) - 5} = \frac{-2}{-5} = \frac{2}{5}$
The curve crosses the $y$-axis at $\left(0, \frac{2}{5}\right)$.

3. Find the equations of the asymptotes.

The vertical asymptote occurs where the denominator is zero (which corresponds to $t = \frac{1}{2}$ in parametric form):
$2x - 5 = 0 \implies x = \frac{5}{2}$

The horizontal asymptote is found by taking the limit as $x \to \pm\infty$ (or $t \to \pm\infty$):
$y \to \frac{x}{2x} = \frac{1}{2}$
The equations of the asymptotes are $x = \frac{5}{2}$ and $y = \frac{1}{2}$.

4. Sketch the curve using the found features.

Draw the coordinate axes and indicate the calculated features:

• Draw vertical dotted line at $x = \frac{5}{2}$ and horizontal dotted line at $y = \frac{1}{2}$.
• Plot and label the points $(2, 0)$ and $\left(0, \frac{2}{5}\right)$.
• Draw the two branches of the hyperbola, with the lower-left branch passing through the intercept points and approaching the asymptotes, and the upper-right branch symmetrically placed in the $x > \frac{5}{2}$, $y > \frac{1}{2}$ region.

1. Find the gradient function $\frac{dy}{dx}$ in terms of $t$ using the parametric equations.

Given $x = t + 2$ and $y = \dfrac{t}{2t - 1}$.

Differentiating both with respect to $t$:
$\dfrac{dx}{dt} = 1$

Using the quotient rule for $y$:
$\dfrac{dy}{dt} = \dfrac{1(2t - 1) - t(2)}{(2t - 1)^2} = \dfrac{-1}{(2t - 1)^2}$

Therefore, $\dfrac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} = \dfrac{-1}{(2t - 1)^2}$

2. Formulate the equation of the tangent at $t = p$.

At $t = p$, the gradient of the tangent is $m = \dfrac{-1}{(2p - 1)^2}$.
The coordinates of the point on the curve are $\left(p+2, \dfrac{p}{2p-1}\right)$.

Using the point-gradient form $y - y_1 = m(x - x_1)$, the equation of the tangent is:
$y - \dfrac{p}{2p-1} = \dfrac{-1}{(2p-1)^2} \left(x - (p+2)\right)$

3. Find the $x$-coordinate of the point A where the tangent intersects the $x$-axis.

At the $x$-axis, $y = 0$:
$0 - \dfrac{p}{2p-1} = \dfrac{-1}{(2p-1)^2}(x - p - 2)$

Multiplying both sides by $-(2p-1)^2$:
$p(2p-1) = x - p - 2$
$2p^2 - p = x - p - 2$
$x = 2p^2 + 2$

Thus, the coordinates of point A are $(2p^2 + 2, 0)$.

4. Find the $y$-coordinate of the point B where the tangent intersects the $y$-axis.

At the $y$-axis, $x = 0$:
$y - \dfrac{p}{2p-1} = \dfrac{-1}{(2p-1)^2}(0 - p - 2)$
$y = \dfrac{p}{2p-1} + \dfrac{p+2}{(2p-1)^2}$

To combine into a single fraction, express with a common denominator of $(2p-1)^2$:
$y = \dfrac{p(2p-1) + p + 2}{(2p-1)^2}$
$y = \dfrac{2p^2 - p + p + 2}{(2p-1)^2} = \dfrac{2p^2 + 2}{(2p-1)^2}$

Thus, the coordinates of point B are $\left(0, \dfrac{2p^2 + 2}{(2p-1)^2}\right)$.

5. Calculate the area of triangle OAB.

Since $2p^2 + 2 = 2(p^2 + 1) > 0$ for all real $p$, and $(2p-1)^2 > 0$ given $p \ne \frac{1}{2}$, both intercept values are positive.

Area of triangle OAB = $\dfrac{1}{2} \times OA \times OB$
$= \dfrac{1}{2}(2p^2 + 2)\left(\dfrac{2p^2 + 2}{(2p-1)^2}\right)$
$= \dfrac{1}{2} \cdot \dfrac{4(p^2 + 1)^2}{(2p-1)^2}$
$= \dfrac{2(p^2 + 1)^2}{(2p-1)^2}$