The curve C has equation y = 2x - 8 times ext{ } oyalblue{ ext{ }} ext{ }rac{1}{2} imes oyalblue{ ext{ }} ext{ } imes x + 5, ext{ }x > 0 a) Find \frac{dy}{dx}, giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 3

For the point P where x = \frac{1}{4}:

1. Calculate \frac{dy}{dx}\ at P: $\frac{dy}{dx}\Big|_{x=\frac{1}{4}} = 2 - \frac{4}{\sqrt{\frac{1}{4}}} = 2 - 8 = -6$

2. Find the y-coordinate of point P: $y = 2\cdot \frac{1}{4} - 8\sqrt{\frac{1}{4}} + 5 = \frac{1}{2} - 4 + 5 = 2.5$

3. Now using point-slope form of the line, we use the coordinates (\frac{1}{4}, 2.5) and slope -6: $y - 2.5 = -6\left(x - \frac{1}{4}\right)$

   Rearranging gives: $y = -6x + 2.5 + \frac{3}{2} = -6x + 6$ Thus, in the form y = ax + b, we find: $y = -6x + 6$

Since the tangent at Q is parallel to the given line 2x - 3y + 18 = 0, we first determine its slope:

Rearranging gives: $3y = 2x + 18 \implies y = \frac{2}{3}x + 6$

The slope is \frac{2}{3}.

Now we set \frac{dy}{dx} equal to this slope: $2 - \frac{4}{\sqrt{x}} = \frac{2}{3}$ Rearranging and simplifying: $-\frac{4}{\sqrt{x}} = \frac{2}{3} - 2$ $-\frac{4}{\sqrt{x}} = \frac{2 - 6}{3} = -\frac{4}{3}$

Thus, we cross-multiply: $4\cdot 3 = 4\sqrt{x}$ $\sqrt{x} = 3 \implies x = 9$

Now, we find the corresponding y-coordinate using the curve equation: $y = 2\cdot 9 - 8\sqrt{9} + 5 = 18 - 24 + 5 = -1$

So, the coordinates of Q are (9, -1).