Figure 2 shows a sketch of the curve C with parametric equations $$x = 1 + t - 5 \sin(t),$$ $$y = 2 - 4 \, ext{cost},$$ $$-\pi < t < \pi$$ The point A lies on the curve C - Edexcel - A-Level Maths Pure - Question 6 - 2018 - Paper 9

To find the equation of the tangent line at point A(k, 2), we first need to calculate the derivatives of x and y with respect to t.

From the given parametric equations:

$x = 1 + t - 5\sin(t)$ $y = 2 - 4\cos(t)$

Next, we find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

$\frac{dx}{dt} = 1 - 5\cos(t)$ $\frac{dy}{dt} = 4\sin(t)$

The slope of the tangent line is given by:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4\sin(t)}{1 - 5\cos(t)}$

At the point A, we can substitute (t = \frac{\pi}{2}) to find the slope:

$\frac{dy}{dx} = \frac{4\sin(\frac{\pi}{2})}{1 - 5\cos(\frac{\pi}{2})} = \frac{4}{1 - 0} = 4$

Thus, the slope of the tangent at A is 4. Using point-slope form,

$y - y_1 = m(x - x_1)$,

where (m = 4), (y_1 = 2), and (x_1 = 6 - \frac{\pi}{2}):

$y - 2 = 4\left(x - \left(6 - \frac{\pi}{2}\right)\right)$

Simplifying:

$y - 2 = 4x - 24 + 2\pi$

Thus,

$y = 4x - 22 + 2\pi$

The final tangent equation in the form y = px + q is:

$y = 4x + (2\pi - 22)$

where p = 4 and q = (2π - 22).