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 - 2008 - Paper 7

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

• For x: $\frac{dx}{dt} = 3t^2 - 8$
• For y: $\frac{dy}{dt} = 2t$

At t = -1:

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

The gradient m(T) is:

$m(T) = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2}{-5} = \frac{2}{5}$

Using point-slope form, the equation of the tangent line l at point A (7, 1) is:

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

This simplifies to:

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

Thus, the equation of line l is confirmed as $2x - 5y - 9 = 0$.

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

1. Substitute $x$ and $y$: $y = \frac{2}{5}x + \frac{9}{5}$

2. Now plug in the parametric equations: $y = t^2$ $x = t^3 - 8t$

We equate $y$ from both equations: $t^2 = \frac{2}{5}(t^3 - 8t) + \frac{9}{5}$ Multiply through by 5 to eliminate the fraction: $5t^2 = 2(t^3 - 8t) + 9$ So: $5t^2 = 2t^3 - 16t + 9$ Rearranging gives: $2t^3 - 5t^2 - 16t + 9 = 0$

3. By synthetic division or other methods, find the roots. One of the roots is t = 3.

4. Substitute t = 3 into the parametric equations: $x = 3^3 - 8(3) = 27 - 24 = 3$ $y = 3^2 = 9$

Thus, the coordinates of point B are (3, 9).