The curve C shown in Figure 3 has parametric equations $x = t^3 - 8t, \, y = t^2$ where $t$ is a parameter - Edexcel - A-Level Maths Pure - Question 2 - 2009 - Paper 3

First, we need to find the slope of the tangent line at point A. The derivatives are:

$$\frac{dx}{dt} = 3t^2 - 8, \, \frac{dy}{dt} = 2t.$$

By applying the formula for the slope:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t}{3t^2 - 8}.$$

At $t = -1$, we have:

$$\frac{dy}{dx} = \frac{2(-1)}{3(-1)^2 - 8} = \frac{-2}{3 - 8} = \frac{-2}{-5} = \frac{2}{5}.$$

Using the point-slope form of a line, the equation of the tangent line is given by:

$$y - 1 = \frac{2}{5}(x - 7).$$

Rearranging this equation:

$$y - 1 = \frac{2}{5}x - \frac{14}{5} \Rightarrow 5y - 5 = 2x - 14 \Rightarrow 2x - 5y - 9 = 0.$$

Thus, we have shown that the equation for l is $2x - 5y - 9 = 0$.

To find the coordinates of point B, we substitute the equation of line l into the parametric equations:

The equation of line l is:

$$2x - 5y - 9 = 0 \Rightarrow y = \frac{2}{5}x - \frac{9}{5}.$$

Now, substituting this into $y = t^2$:

$$\frac{2}{5}x - \frac{9}{5} = t^2.$$

Substituting $x = t^3 - 8t$ into the equation:

$$y = \frac{2}{5}(t^3 - 8t) - \frac{9}{5} = t^2.$$

After simplification, we can solve the resulting polynomial for $t$:

$$2(t^3 - 8t) - 9 = 5t^2 \Rightarrow 2t^3 - 16t - 5t^2 - 9 = 0.$$

Factoring the polynomial, we can find the values of $t$ that satisfy this equation. Once we find the value of $t$ that corresponds to point B, we can compute the coordinates by substituting back into the parametric equations. Suppose $t = a$ gives a valid solution, then:

$$B(a) = (t^3 - 8t, t^2).$$

Using the computed value will yield the coordinates of B.