Differentiate \( f(x) = x^2 – 10 \times x \):
\( f'(x) = \frac{d}{dx}(x^2 – 10 \times x) = 2 \times x – 10 \) (M1).

At \( x = 4 \): \( f'(4) = 2 \times 4 – 10 = 8 – 10 = -2 \).
Gradient of tangent at \( A \): \( -2 \) (A1 N2).

[2 marks]

Gradient of normal is the negative reciprocal of the tangent’s gradient:
\( -\frac{1}{-2} = \frac{1}{2} \) (A1 N1).

[1 mark]

Use point-slope form at \( A(4, -24) \) with gradient \( \frac{1}{2} \):
\( y – (-24) = \frac{1}{2} \times (x – 4) \).
Simplify: \( y + 24 = \frac{1}{2} \times x – 2 \implies y = \frac{1}{2} \times x – 26 \).
Answer: \( y = \frac{1}{2} \times x – 26 \) (A1 N1).

[1 mark]

Method 1: Quadratic Formula
Solve for intersection of normal \( y = \frac{1}{2} \times x – 26 \) and curve \( f(x) = x^2 – 10 \times x \):
\( x^2 – 10 \times x = \frac{1}{2} \times x – 26 \) (M1).

Multiply by 2: \( 2 \times x^2 – 20 \times x = x – 52 \).
Simplify: \( 2 \times x^2 – 21 \times x + 52 = 0 \) (A1).

Quadratic formula: \( x = \frac{21 \pm \sqrt{(-21)^2 – 4 \times 2 \times 52}}{2 \times 2} = \frac{21 \pm \sqrt{441 – 416}}{4} = \frac{21 \pm \sqrt{25}}{4} \).
\( x = \frac{21 \pm 5}{4} \implies x = \frac{26}{4} = 6.5, x = \frac{16}{4} = 4 \).

\( x = 4 \) is point \( A \), so \( x = 6.5 \) is point \( B \).
Find \( y \): \( y = \frac{1}{2} \times 6.5 – 26 = 3.25 – 26 = -22.75 \).
Verify: \( f(6.5) = 6.5^2 – 10 \times 6.5 = 42.25 – 65 = -22.75 \).
Coordinates of \( B \): \( (6.5, -22.75) \) (A1 N2).

Method 2: Completing the Square
Rewrite: \( 2 \times x^2 – 21 \times x + 52 = 0 \).
Divide by 2: \( x^2 – \frac{21}{2} \times x + 26 = 0 \).
Complete the square: \( x^2 – \frac{21}{2} \times x = -\frac{52}{2} \implies \left(x – \frac{21}{4}\right)^2 – \left(\frac{21}{4}\right)^2 = -26 \).
\( \left(x – \frac{21}{4}\right)^2 = \frac{441}{16} – \frac{416}{16} = \frac{25}{16} \).
\( x – \frac{21}{4} = \pm \frac{5}{4} \implies x = \frac{21 \pm 5}{4} = 6.5, 4 \) (M1 A1).
\( x = 6.5 \), \( y = -22.75 \) (as above). Coordinates of \( B \): \( (6.5, -22.75) \) (A1 N2).

[3 marks]