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

We start by differentiating:

1. Given $y = 2t$, the derivative with respect to $t$ is: $\frac{dy}{dt} = 2$

2. For $x = t^2$, we have: $\frac{dx}{dt} = 2t$

3. The gradient of the curve (slope) is: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{2t} = \frac{1}{t}$

4. At point A, where $t = a$, the gradient becomes: $\frac{dy}{dx} \bigg|_{t=a} = \frac{1}{a}$

5. Therefore, we have: $\tan \theta = \frac{1}{a}$

Considering points A and B:

1. Point A has coordinates $(a^2, 2a)$ and point B has coordinates $(1, 0)$.

2. The gradient of line AB is: $\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 2a}{1 - a^2} = \frac{-2a}{1 - a^2}$

3. Therefore, the tangent of angle $\\phi$ can be expressed as: $\tan \phi = \frac{-2a}{1 - a^2}$

Using the double angle formula for tangent:

1. We have: $\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$

2. Substituting in $an \theta = \frac{1}{a}$: $\tan 2\theta = \frac{2(\frac{1}{a})}{1 - (\frac{1}{a})^2} = \frac{\frac{2}{a}}{\frac{a^2 - 1}{a^2}} = \frac{2a}{a^2 - 1}$

3. Earlier, we found:\n $\tan \phi = \frac{-2a}{1 - a^2} = \frac{2a}{a^2 - 1}$

Thus: $\tan 2\theta = \tan \phi$