2026-08-31T11:35:06.667100 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 4 shows part of the curve \(C\) with equat...

(a) \(\dfrac{dy}{dx}\) and the gradient at \(P\) [2]

The gradient of a curve at a point is the value of its derivative there, so differentiate \(y = x^{2} - 6x + 13\) and substitute \(x = 4\).

\[\frac{dy}{dx} = 2x - 6 \quad \textbf{[M1]}\] \[\text{At } x = 4: \quad \frac{dy}{dx} = 8 - 6 = 2,\]

so the gradient of \(C\) at \(P\) is \(2\). [A1] The M mark for a correct derivative survives an evaluation slip; the A mark is for the value \(2\).

(b) Show that the tangent at \(P\) has equation \(y = 2x - 3\) [3]

A straight line needs a gradient and a point, so first find the \(y\) coordinate of \(P\) from the equation of the curve:

\[\text{At } x = 4: \quad y = 16 - 24 + 13 = 5, \quad \text{so } P(4,\, 5). \quad \textbf{[B1]}\]

This is a B mark, given on sight for the correct point. Then use \(y - y_{1} = m(x - x_{1})\) with \(m = 2\):

\[y - 5 = 2(x - 4) \quad \textbf{[M1]}\] \[y = 2x - 8 + 5 = 2x - 3 \quad \textbf{[A1]}\]

The equation is printed in the question, so the marks are for the derivation. Do not substitute \(x = 4\) into the derivative to get the \(y\) coordinate: the derivative gives the gradient, the curve gives the height.

(c) Equation of the normal at \(P\) in the form \(ax + by + c = 0\) [3]

The normal is perpendicular to the tangent, and perpendicular gradients multiply to \(-1\), so the normal has gradient

\[m_{n} = -\frac{1}{2} \quad \textbf{[M1]}\] \[y - 5 = -\frac{1}{2}(x - 4) \quad \textbf{[M1]}\]

Multiply through by \(2\) to clear the fraction and collect everything on one side, because the question demands integer coefficients:

\[2y - 10 = -(x - 4) = -x + 4 \quad \Rightarrow \quad x + 2y - 14 = 0 \quad \textbf{[A1]}\]

Both method marks are independent of the arithmetic; the accuracy mark needs the required form with integers, so \(y = -0.5x + 7\) does not satisfy the instruction.

(d) Area of triangle \(ABP\) [2]

The two lines cross the \(y\) axis at \(A\) and \(B\), so put \(x = 0\) in each equation:

• Tangent: \(y = 2(0) - 3 = -3\), so \(A(0,\, -3)\).
• Normal: \(0 + 2y - 14 = 0 \Rightarrow y = 7\), so \(B(0,\, 7)\).

Because \(A\) and \(B\) both lie on the \(y\) axis, \(AB\) is a vertical segment and is the natural base:

\[AB = 7 - (-3) = 10.\]

The corresponding perpendicular height is the horizontal distance from \(P\) to the \(y\) axis, which is simply the \(x\) coordinate of \(P\), namely \(4\). [M1]

\[\text{Area} = \frac{1}{2} \times 10 \times 4 = 20 \ \text{square units} \quad \textbf{[A1]}\]

Common wrong turns on this question. Taking the gradient of the normal as \(-2\) or as \(\dfrac{1}{2}\) rather than the negative reciprocal \(-\dfrac{1}{2}\). Computing \(AB\) as \(7 + (-3) = 4\) instead of subtracting the coordinates. In part (d), using the length \(AP\) or \(BP\) as the height: the height must be measured perpendicular to the chosen base, and here the base is vertical, so the height is horizontal and equals \(4\). Using \(\dfrac{1}{2}ab\sin C\) with the angle at \(P\) also works but is far more effort, and the right angle at \(P\) gives a third valid route: \(AP = \sqrt{16 + 64} = 4\sqrt{5}\), \(BP = \sqrt{16 + 4} = 2\sqrt{5}\), and \(\dfrac{1}{2} \times 4\sqrt{5} \times 2\sqrt{5} = 20\).

Check. That third route agreeing with the base-and-height calculation is the strongest available check, and it also confirms the perpendicularity: the tangent and normal meet at right angles at \(P\), so \(AP\) and \(BP\) are the two legs of a right-angled triangle. A quick reading off the figure also confirms \(A\) below the origin and \(B\) above it.