Figure 2 shows a sketch of the curve $H$ with equation $y = \frac{3}{x} + 4, \ x \neq 0.$ (a) Give the coordinates of the point where $H$ crosses the $x$-axis - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 1

The asymptotes for the curve can be determined by analyzing the behavior as $x$ approaches $0$ and infinity.

1. Vertical asymptote: As $x \to 0$, $y$ approaches infinity, so the equation is: $x = 0$ (the $y$-axis).
2. Horizontal asymptote: As $x \to \infty$ or $x \to -\infty$, $y$ approaches $4$, hence: $y = 4$.

First, we find the gradient of the curve at point $P(-3, 3)$ by differentiating:

$y = \frac{3}{x} + 4$

Thus,

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

At $x = -3$:

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

The gradient of the normal is the negative reciprocal of the tangent gradient:

$\text{Normal Gradient} = -\frac{1}{(-\frac{1}{3})} = 3$

Using the point-slope form of a line equation:

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

we substitute $(-3, 3)$ and the gradient:

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

This simplifies to:

$y = 3x + 12$.

To find points $A$ and $B$, we determine where the normal intersects the axes.

1. For $A$, set $y = 0$ in the normal equation:

   $0 = 3x + 12 \Rightarrow 3x = -12 \Rightarrow x = -4$

   Thus, $A(-4, 0)$.

2. For $B$, set $x = 0$:

   $y = 3(0) + 12 = 12 \Rightarrow B(0, 12).$

Now, use the distance formula to find $AB$:

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(0 - (-4))^2 + (12 - 0)^2} = \sqrt{(4)^2 + (12)^2} = \sqrt{16 + 144} = \sqrt{160} = 4\sqrt{10}.$