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 1 - 2012 - Paper 1

First, we need to find the derivative of the function:

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

At point A, where $x = \frac{1}{2}$, we calculate the gradient:

$\frac{dy}{dx} = 4.$

The gradient of the normal is the negative reciprocal, so:

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

Using the point-slope form of the line:

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

substituting in the coordinates of A:

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

Rearranging this gives the equation of the normal. After simplifying, we arrive at:

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

To find point B, we set the equation of the normal equal to the curve equation:

1. Substitute the normal equation into the curve equation.
2. Solve for the intersection points.

We found previously that the normal at A is given by:

$2x + 8y - 1 = 0$

Substituting (y = 2 - \frac{1}{x}) gives:

$2x + 8(2 - \frac{1}{x}) - 1 = 0$

This simplifies to:

$2x + 16 - \frac{8}{x} - 1 = 0$

Rearranging yields:

$2x^2 + 15x - 8 = 0.$

Using the quadratic formula to solve for x results in:

$x = -\frac{1}{2} \text{ or } x = -4.$

After finding the corresponding y-values for these x-coordinates, we conclude:

• For (x = -\frac{1}{2}), (y = \frac{17}{8}).
• For (x = -4), (y = 2 - \frac{1}{-4} = \frac{9}{4}= 2.25).

Thus, the coordinates of B are ((-4, 2.25)).