Figure 2 shows a sketch of the curve $H$ with equation $y = \frac{3}{x} + 4, \quad x \neq 0$ - Edexcel - A-Level Maths Pure - Question 3 - 2013 - Paper 1

The curve has two asymptotes:

1. Vertical Asymptote: Set the denominator to zero (where $x = 0$): $x = 0$

2. Horizontal Asymptote: For large values of $|x|$, $y$ approaches $4$ as $x \to \pm \infty$: $y = 4$

To find the normal at the point $P(-3, 3)$, we first need to calculate the gradient of the curve at this point.

Using the derivative:

$\frac{dy}{dx} = -\frac{3}{x^2}$

Evaluating at $x = -3$:

$\frac{dy}{dx} = -\frac{3}{(-3)^2} = -\frac{3}{9} = -\frac{1}{3}$

The gradient of the normal is the negative reciprocal:

$m_{normal} = -\frac{1}{m} = 3$

Using the point-slope form of the equation $y - y_1 = m(x - x_1)$, where $m = 3$ and $(x_1, y_1) = (-3, 3)$:

$y - 3 = 3(x + 3)$

Simplifying gives:

$y = 3x + 12$

The endpoints of segment $AB$ are $A(-4, 0)$ and $B(0, 12)$. Using the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting in our values:

$d = \sqrt{(0 - (-4))^2 + (12 - 0)^2}$ $= \sqrt{(4)^2 + (12)^2}$ $= \sqrt{16 + 144}$ $= \sqrt{160}$ $= 4\sqrt{10}$

Thus, the length of segment $AB$ is $4\sqrt{10}$.