(a)

(i)

\( f'(x) = -k x^{-2} \)

\( f'(p) = -k p^{-2} = -\frac{k}{p^2} \). A1

[2 marks]

(ii)

Attempt to use point and gradient to find equation of \( L_1 \). M1

e.g., \( y – \frac{k}{p} = -k p^{-2} (x – p) \), or \( \frac{k}{p} = -\frac{k}{p^2} p + b \)

Correct working leading to answer:

e.g., \( p^2 y – k p = -k x + k p \), or \( y – \frac{k}{p} = -\frac{k}{p^2} x + \frac{2k}{p} \). M1

\( k x + p^2 y – 2p k = 0 \). A1

[4 marks]

(b)

METHOD 1

Area of a triangle. M1

Recognizing \( x = 0 \) at \( B \). M1

Correct working to find \( y \)-coordinate of \( B \):

\( p^2 y – 2p k = 0 \implies y = \frac{2k}{p} \) (may be seen in area formula). A1

Correct substitution to find area of triangle:

e.g., \( \frac{1}{2} (2p) \left( \frac{2k}{p} \right) \), or \( p \times \frac{2k}{p} \). M1

Area of triangle \( AOB = 2k \). A1

METHOD 2

Integration, recognizing to integrate \( L_1 \) between 0 and \( 2p \). M1

e.g., \( \int_0^{2p} L_1 \, dx \), or \( \int_0^{2p} \left( -\frac{k}{p^2} x + \frac{2k}{p} \right) \, dx \)

Correct integration of both terms:

e.g., \( -\frac{k x^2}{2 p^2} + \frac{2k x}{p} \). M1

Substituting limits into integrated function and subtracting:

e.g., \( \left( -\frac{k (2p)^2}{2 p^2} + \frac{2k (2p)}{p} \right) – (0) \), or \( -\frac{4k p^2}{2 p^2} + \frac{4k p}{p} \). M1

Correct working: e.g., \( -2k + 4k \). A1

Area of triangle \( AOB = 2k \). A1

[5 marks]

(c)

Recognizing use of transformation. M1

e.g., area of triangle \( AOB = \) area of triangle \( DEF \), \( g(x) = \frac{k}{x – 4} + 3 \), gradient of \( L_2 = \) gradient of \( L_1 \), \( D(4, 3) \), \( 2p + 4 \), one correct shift.

Correct working:

e.g., area of triangle \( DEF = 2k \), \( CD = 3 \), \( DF = 2p \), \( CG = 2p \), \( E \left( 4, \frac{2k}{p} + 3 \right) \). A1

Area of rectangle \( CDFG = 2k \). A1

Valid approach:

e.g., \( \frac{ED \times DF}{2} = CD \times DF \), or \( 2p \cdot 3 = 2k \), or \( ED = CD \), or \( \int_4^{2p + 4} L_2 \, dx = 4k \). M1

Correct working:

e.g., \( ED = 6 \), \( E(4, 9) \), \( k = 3p \), gradient = \( \frac{3 – \left( \frac{2k}{p} + 3 \right)}{(2p + 4) – 4} \), or \( \frac{-6}{\frac{2k}{3}} \), or \( -\frac{9}{k} \). M1

e.g., \( \frac{-6}{2p} \), or \( \frac{9 – 3}{(2p + 4) – 4} \), or \( -\frac{3p}{p^2} \), or \( \frac{3 – \left( \frac{2(3p)}{p} + 3 \right)}{(2p + 4) – 4} \), or \( -\frac{9}{3p} \). A1

Gradient of \( L_2 \) is \( -\frac{3}{p} = -3 p^{-1} \). A1

[6 marks]