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

To find the gradient of the curve at the point A where $t = a$, we differentiate:

1. The equation $x = t^2$ gives us:

   $\frac{dx}{dt} = 2t\text{, hence at } t = a: \frac{dx}{dt} = 2a$

2. The equation $y = 2t$ yields:

   $\frac{dy}{dt} = 2\text{, hence at } t = a: \frac{dy}{dt} = 2$

3. Now, we can find the gradient of the tangent line:

   $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{2a} = \frac{1}{a}$

4. The tangent of the angle $\theta$ between the tangent and the x-axis is equal to the gradient:

   $\tan \theta = \frac{1}{a}$

The line AB is vertical since point A is on the curve. Therefore, the coordinates of A can be given as $(x_A, y_A)$ or $(a^2, 2a)$. The slope from point A to point B is given by:

1. The coordinates of point B are $(1, 0)$.

2. The gradient $m$ of line AB is:

   $m = \frac{y_B - y_A}{x_B - x_A} = \frac{0 - 2a}{1 - a^2} = \frac{-2a}{1 - a^2}$

3. Thus, we have:

   $\tan \phi = \frac{-2a}{1 - a^2}$

To show that $\tan 2\theta = \tan \phi$, we use the double angle formula:

1. The formula states:

   $\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$

2. Substituting $\tan \theta = \frac{1}{a}$, we find:

   $\tan 2\theta = \frac{2(\frac{1}{a})}{1 - (\frac{1}{a})^2} = \frac{\frac{2}{a}}{1 - \frac{1}{a^2}} = \frac{2}{a} \cdot \frac{a^2}{a^2 - 1} = \frac{2a}{a^2 - 1}$

3. Since we earlier found $\tan \phi = \frac{-2a}{1 - a^2}$, we observe:

   $\tan \phi = \frac{2a}{a^2 - 1}$

Thus, we conclude:

$\tan 2\theta = \tan \phi$