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

The curve has two asymptotes:

1. Vertical Asymptote: This occurs where the denominator of the equation approaches zero, i.e., when $x = 0$.

2. Horizontal Asymptote: As $x$ approaches infinity (both positive and negative), the term $\frac{3}{x}$ approaches 0, leading to the asymptote:

$y = 4$.

To find the normal at point P, we first need the derivative of $y$:

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

Differentiating:

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

At $x = -3$:

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

The slope of the normal line is the negative reciprocal:

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

Using point-slope form, the equation of the normal line at P(-3, 3) is given by:

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

Thus,

$y = 3x + 12.$

To find the coordinates of A and B:

• The x-intercept (A) occurs when $y = 0$: From the normal equation $y = 3x + 12$, setting $y = 0$: $0 = 3x + 12$ $x = -4$ So, $A(-4, 0)$.

• The y-intercept (B) occurs when $x = 0$: $y = 3(0) + 12 = 12$ So, $B(0, 12)$.

Now, using the distance formula: $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for $A(-4, 0)$ and $B(0, 12)$: $AB = \sqrt{(0 - (-4))^2 + (12 - 0)^2}$ $= \sqrt{(4)^2 + (12)^2}$ $= \sqrt{16 + 144}$ $= \sqrt{160}$ $= 4\sqrt{10}.$