The curve defined by the parametric equations $x = t^2$ and $y = 2t$ is shown in Figure 1 below - AQA - A-Level Maths Pure - Question 8 - 2020 - Paper 2

First, we differentiate the parametric equations
$x = t^2 \Rightarrow \frac{dx}{dt} = 2t$
and
$y = 2t \Rightarrow \frac{dy}{dt} = 2.$
Using the formula for the gradient of the curve, we can find
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{2t} = \frac{1}{t}.$
At the point A where $t = a$, we have:
$$\tan \theta = \frac{dy}{dx} = \frac{1}{a}.$

The gradient of line AB, which connects points A and B, where A has coordinates $(t^2, 2t)$ and B is $(1, 0)$, is given by:
$\tan \phi = \frac{y_B - y_A}{x_B - x_A} = \frac{0 - 2a}{1 - a^2} = \frac{-2a}{1 - a^2}.$

Using the double angle formula for tangent, we have:
$\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}.$
Substituting $\tan \theta = \frac{1}{a}$ gives:
$\tan 2\theta = \frac{2(\frac{1}{a})}{1 - (\frac{1}{a})^2} = \frac{\frac{2}{a}}{1 - \frac{1}{a^2}} = \frac{\frac{2}{a}}{\frac{a^2 - 1}{a^2}} = \frac{2a}{a^2 - 1}.$
Now, equating this to $\tan \phi = \frac{-2a}{1 - a^2}$, we see that both expressions are equal, demonstrating that $\tan 2\theta = \tan \phi$.