2026-08-31T11:35:09.827994 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 5 shows the curve \(C\) with equation \(y ...

(a) Equations of the two asymptotes of \(C\) [2]

An asymptote is a line the curve approaches without ever meeting. For \(y = \dfrac{12}{x}\) there are two:

• As \(x\) approaches \(0\) the value of \(\dfrac{12}{x}\) grows without bound, so the vertical asymptote is \(x = 0\), the \(y\) axis. [B1]
• As \(x\) grows large in either direction \(\dfrac{12}{x}\) approaches \(0\), so the horizontal asymptote is \(y = 0\), the \(x\) axis. [B1]

Two independent B marks. Both answers must be given as equations of lines: writing "the \(x\) axis and the \(y\) axis" in words, or writing \(0\) alone, does not satisfy "write down the equations". The exclusion \(x \neq 0\) in the question is the hint for the vertical one.

(b) Show that the \(x\) coordinates of \(A\) and \(B\) satisfy \(x^{2} - 8x + 12 = 0\) [2]

At a point of intersection the two \(y\) values agree, so equate the equations of \(C\) and \(l\):

\[\frac{12}{x} = 8 - x \quad \textbf{[M1]}\]

Clear the fraction by multiplying every term by \(x\). This is legitimate because \(x \neq 0\) on the curve, and saying so is good practice since multiplying by a possibly zero quantity can introduce spurious roots:

\[12 = 8x - x^{2} \quad \Rightarrow \quad x^{2} - 8x + 12 = 0 \quad \textbf{[A1]}\]

The target is printed, so both marks are for the derivation and quoting the equation earns nothing. Multiplying only the left side by \(x\), or dropping the \(x\) from \(8x\), are the slips to guard against.

(c) Coordinates of \(A\) and of \(B\) [2]

The quadratic factorises, since \(-2\) and \(-6\) multiply to \(12\) and add to \(-8\):

\[(x - 2)(x - 6) = 0 \quad \Rightarrow \quad x = 2 \ \text{ or } \ x = 6 \quad \textbf{[M1]}\]

Find each \(y\) from either equation, the line being the simpler: at \(x = 2\), \(y = 8 - 2 = 6\); at \(x = 6\), \(y = 8 - 6 = 2\). The question states that \(A\) has the smaller \(x\) coordinate, so

\[A(2,\, 6) \qquad \text{and} \qquad B(6,\, 2) \quad \textbf{[A1]}\]

Assigning the labels the other way round is the error the question's own wording is there to prevent, and the accuracy mark depends on getting them the right way about.

(d) Exact length of \(AB\) in the form \(k\sqrt{2}\) [3]

Use the distance formula, which is Pythagoras applied to the horizontal and vertical separations:

\[AB^{2} = (6 - 2)^{2} + (2 - 6)^{2} \quad \textbf{[M1]}\] \[AB^{2} = 4^{2} + (-4)^{2} = 16 + 16 = 32 \quad \textbf{[M1]}\]

Now simplify the surd by extracting the largest square factor, \(32 = 16 \times 2\):

\[AB = \sqrt{32} = \sqrt{16}\sqrt{2} = 4\sqrt{2} \quad \textbf{[A1]}\]

The two method marks survive arithmetic slips; the accuracy mark requires the exact surd form, so \(5.66\) does not answer the question as set. Note that squaring removes the sign, so \((2 - 6)^{2} = 16\), not \(-16\); a negative contribution under the root is a sure sign of an error.

Checks. Both points must lie on the curve as well as the line: \(\dfrac{12}{2} = 6\) and \(\dfrac{12}{6} = 2\), so \(A(2,\,6)\) and \(B(6,\,2)\) are correct. The symmetry is a further check: \(A\) and \(B\) are reflections of each other in the line \(y = x\), which is expected because both \(y = \dfrac{12}{x}\) and \(y = 8 - x\) are unchanged when \(x\) and \(y\) are swapped. Finally, since the line \(l\) has gradient \(-1\), the horizontal and vertical separations between \(A\) and \(B\) must be equal, and both are \(4\), which is exactly why the answer comes out as a multiple of \(\sqrt{2}\).