Figure 2 shows a sketch of the curve C with equation y = 2 - \frac{1}{x}, \quad x \neq 0 The curve crosses the x-axis at the point A: (a) Find the coordinates of A - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 2

The gradient of the curve at A can be found by differentiating the curve equation:

$y = 2 - \frac{1}{x}$

Differentiating yields:

$\frac{dy}{dx} = \frac{1}{x^2}$

At point A where $x = \frac{1}{2}$:

$\frac{dy}{dx} \Bigg|_{x = \frac{1}{2}} = 4$

The gradient of the normal line is the negative reciprocal of the tangent's gradient:

$m = -\frac{1}{4}$

Using the point-slope form of the line equation:

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

Substituting in the coordinates of A:

$y - 0 = -\frac{1}{4}(x - \frac{1}{2})$

This simplifies to:

$y = -\frac{1}{4}x + \frac{1}{8}$

To express this in standard form:

$\rightarrow 4y + x - \frac{1}{2} = 0$

Multiplying through by 2 gives:

$2x + 8y - 1 = 0$.

To find the coordinates of point B, we need to substitute the normal line equation into the curve equation:

1. The equation of the normal line: $2x + 8y - 1 = 0$ can be rearranged to

$y = -\frac{1}{4}x + \frac{1}{8}$.

2. Set this equal to the curve equation:

$2 - \frac{1}{x} = -\frac{1}{4}x + \frac{1}{8}$.

3. Cross-multiplying yields:

$2x - 1 = -\frac{1}{4}x^2 + \frac{1}{8}x$

4. Finally, after simplifying and solving the resulting quadratic equation, we can obtain the coordinates of B, which will give:

$B(-8, 1)$.