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

1. Differentiate $y = 2t$ with respect to $t$: $\frac{dy}{dt} = 2.$

2. Also differentiate $x = t^2$ with respect to $t$: $\frac{dx}{dt} = 2t.$

3. The gradient of the curve rac{dy}{dx} is given by: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{2t} = \frac{1}{t}.$

4. At the point where $t = a$, this gives: $\frac{dy}{dx}|_{t=a} = \frac{1}{a}.$

5. This means the gradient of the tangent line at point A is: $$\tan \theta = \frac{1}{a}.$

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

2. The line AB has a change in $y$ of $2a$ (from point A to the $y$ coordinate of A which is $2a$) and a change in $x$ of $(1 - a^2)$ (from $x = a^2$ to $x = 1$).

3. Therefore, the gradient of line AB is: $\text{Gradient of AB} = \frac{2a - 0}{1 - a^2} = \frac{2a}{1 - a^2}.$

4. Thus, we find: $\tan \phi = \frac{2a}{1 - a^2}.$

1. Using the double angle formula, we have: \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta.

2. Substitute $\tan \theta = \frac{1}{a}$: $\tan 2\theta = \frac{2 \cdot \frac{1}{a}}{1 - \left(\frac{1}{a}\right)^2} = \frac{\frac{2}{a}}{1 - \frac{1}{a^2}} = \frac{\frac{2}{a}}{\frac{a^2 - 1}{a^2}} = \frac{2a}{a^2 - 1}.$

3. Now, we see that: $\tan \phi = \frac{2a}{1 - a^2}$ which simplifies to: $\tan 2\theta = \tan \phi.$

Thus, the statement is proven.