Write down the equation of \({L_1}\).

A line \({L_a}\) crosses the \(y\)-axis at a point \(P\).

Show that \(P\) has coordinates \(\left( {0,{\text{ }}\frac{4}{a}} \right)\).

The line \({L_a}\) crosses the \(x\)-axis at \({\text{Q}}(2a,{\text{ }}0)\). Let \(d = {\text{P}}{{\text{Q}}^2}\).

Show that \(d = 4{a^2} + \frac{{16}}{{{a^2}}}\).

There is a minimum value for \(d\). Find the value of \(a\) that gives this minimum value.

attempt to substitute \(x = 1\)     (M1)

eg\(\;\;\;\)r \( = \left( {\begin{array}{*{20}{c}} 1 \\ {\frac{2}{1}} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} {{1^2}} \\ { - 2} \end{array}} \right),{\text{ }}{L_1} = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \end{array}} \right)\)

correct equation (vector or Cartesian, but do not accept “\({L_1}\)”)

eg\(\;\;\;\)r \( = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \end{array}} \right),{\text{ }}y =  - 2x + 4\;\;\;\)(must be an equation)     A1     N2

[2 marks]

appropriate approach     (M1)

eg\(\;\;\;\left( {\begin{array}{*{20}{c}} 0 \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} {{a^2}} \\ { - 2} \end{array}} \right)\)

correct equation for \(x\)-coordinate     A1

eg\(\;\;\;0 = a + t{a^2}\)

\(t = \frac{{ - 1}}{a}\)     A1

substituting their parameter to find \(y\)     (M1)

eg\(\;\;\;y = \frac{2}{a} - 2\left( {\frac{{ - 1}}{a}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) - \frac{1}{a}\left( {\begin{array}{*{20}{c}} {{a^2}} \\ { - 2} \end{array}} \right)\)

correct working     A1

eg\(\;\;\;y = \frac{2}{a} + \frac{2}{a},{\text{ }}\left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} a \\ { - \frac{2}{a}} \end{array}} \right)\)

finding correct expression for \(y\)     A1

eg\(\;\;\;y = \frac{4}{a},{\text{ }}\left( {\begin{array}{*{20}{c}} 0 \\ {\frac{4}{a}} \end{array}} \right)\) \({\text{P}}\left( {0,{\text{ }}\frac{4}{a}} \right)\)     AG     N0

[6 marks]

valid approach M1

eg\(\;\;\;\)distance formula, Pythagorean Theorem, \(\overrightarrow {{\text{PQ}}}  = \left( {\begin{array}{*{20}{c}} {2a} \\ { - \frac{4}{a}} \end{array}} \right)\)

correct simplification     A1

eg\(\;\;\;{(2a)^2} + {\left( {\frac{4}{a}} \right)^2}\)

\(d = 4{a^2} + \frac{{16}}{{{a^2}}}\)     AG     N0

[2 marks]

recognizing need to find derivative     (M1)

eg\(\;\;\;d',{\text{ }}d'(a)\)

correct derivative     A2

eg\(\;\;\;8a - \frac{{32}}{{{a^3}}},{\text{ }}8x - \frac{{32}}{{{x^3}}}\)

setting their derivative equal to \(0\)     (M1)

eg\(\;\;\;8a - \frac{{32}}{{{a^3}}} = 0\)

correct working     (A1)

eg\(\;\;\;8a = \frac{{32}}{{{a^3}}},{\text{ }}8{a^4} - 32 = 0\)

working towards solution     (A1)

eg\(\;\;\;{a^4} = 4,{\text{ }}{a^2} = 2,{\text{ }}a =  \pm \sqrt 2 \)

\(a = \sqrt[4]{4}\;\;\;(a = \sqrt 2 )\;\;\;({\text{do not accept }} \pm \sqrt 2 )\)     A1     N3

[7 marks]

Total [17 marks]

In part (a), most candidates correctly substituted 1 for \(x\), although many of them did not earn full marks for their work here, as they wrote their vector equation using \({L_1} = \), not understanding that \({L_1}\) is the name of the line, and not a vector.

Very few candidates answered parts (b) and (c) correctly, often working backwards from the given answer, which is not appropriate in "show that" questions. In these types of questions, candidates are required to clearly show their working and reasoning, which will hopefully lead them to the given answer.

Very few candidates answered parts (b) and (c) correctly, often working backwards from the given answer, which is not appropriate in "show that" questions. In these types of questions, candidates are required to clearly show their working and reasoning, which will hopefully lead them to the given answer.

Fortunately, a good number of candidates recognized the need to find the derivative of the given expression for \(d\) in part (d) of the question, and so were able to earn at least some of the available marks in the final part.