The point P lies on the curve with equation y = 4e^{x-2} - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 5

First, we need to determine the derivative of the curve to find the slope of the tangent. The derivative is:

$\frac{dy}{dx} = 4e^{x-2}.$

Next, we substitute for x to find the slope at point P. From part (a), we know:

$x = \frac{1}{2} (\text{ln}(2) - 1).$

Substituting this into the derivative:

$\frac{dy}{dx} = 4e^{\frac{1}{2} (\text{ln}(2) - 1) - 2} = 16.$

Now using the point-slope form of the line:

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

Where

• $y_1 = 8$,
• $x_1 = \frac{1}{2}(\text{ln}(2) - 1)$,
• $m = 16$.

We can plug in those values:

$y - 8 = 16(x - \frac{1}{2} (\text{ln}(2) - 1))$

Simplifying:

$y - 8 = 16x - 8(\text{ln}(2) - 1)$

Thus:

$y = 16x - 16\frac{\text{ln}(2)}{2} + 8.$