Year: 2017
Paper: 2
Question Number: 5
Course: LFM Pure and Mechanics
Section: Parametric equations
This year's paper was, perhaps, slightly more straightforward than usual, with more helpful guidance offered in some of the questions. Thus the mark required for a "1", a Distinction, was 80 (out of 120), around ten marks higher than that which would customarily be required to be awarded this grade. Nonetheless, a three‐figure mark is still a considerable achievement and, of the 1330 candidates sitting the paper, there were 89 who achieved this. At the other end of the scale, there were over 350 who scored 40 or below, including almost 150 who failed to exceed a total score of 25. As a general strategy for success in a STEP examination, candidates should be looking to find four "good" questions to work at (which may be chosen freely by the candidates from a total of 13 questions overall). It is unfortunately the case that so many low‐scoring candidates flit from one question to another, barely starting each one before moving on. There needs to be a willingness to persevere with a question until a measure of understanding as to the nature of the question's purpose and direction begins to emerge. Many low‐scoring candidates fail to deal with those parts of questions which cover routine mathematical processes ‐ processes that should be standard for an A‐level candidate. The significance of the "rule of four" is that four high‐scoring questions (15‐20 marks apiece) obtains you up to around the total of 70 that is usually required for a "1"; and with a couple of supporting starts to questions, such a total should not be beyond a good candidate who has prepared adequately. This year, significantly more than 10% of candidates failed to score at least half marks on any one question; and, given that Q1 (and often Q2 also) is (are) specifically set to give all candidates the opportunity to secure some marks, this indicates that these candidates are giving up too easily. Mathematics is about more than just getting to correct answers. It is about communicating clearly and precisely. Particularly with "show that" questions, candidates need to distinguish themselves from those who are just tracking back from given results. They should also be aware that convincing themselves is not sufficient, and if they are using a result from 3 pages earlier, they should make this clear in their working. A few specifics: In answers to mechanics questions, clarity of diagrams would have helped many students. If new variables or functions are introduced, it is important that students clearly define them. One area which is very important in STEP but which was very poorly done is dealing with inequalities. Although a wide range of approaches such as perturbation theory were attempted, at STEP level having a good understanding of the basics – such as changing the inequality if multiplying by a negative number – is more than enough. In fact, candidates who used more advanced methods rarely succeeded.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
A curve $C$ is determined by the parametric equations
\[
x=at^2 \, , \; y = 2at\,,
\]
where $a > 0$.
\begin{questionparts}
\item Show that the normal to $C$ at a point $P$, with non-zero parameter $p$, meets $C$ again at a point $N$, with parameter $n$, where
\[ n= - \left( p + \frac{2}{p} \right). \]
\item Show that the distance $\left| PN \right|$ is given by
\[ \vert PN\vert^2 = 16a^2\frac{(p^2+1)^3}{p^4} \]
and that this is minimised when $p^2=2\,$.
\item The point $Q$, with parameter $q$, is the point at which the circle with diameter $PN$ cuts $C$ again. By considering the gradients of $QP$ and $QN$, show that
\[ 2 = p^2-q^2 + \frac{2q}p. \]
Deduce that $\left| PN \right|$ is at its minimum when $Q$ is at the origin.
\end{questionparts}\begin{questionparts}
\item $\,$
\begin{align*}
&& \frac{\d x}{\d t} &= 2at \\
&& \frac{\d y}{\d t} &= 2a \\
\Rightarrow && \frac{\d y}{\d x} &= \frac1t \\
&& -p &= \text{grad of normal} \\
&&&= \frac{y-2ap}{x-ap^2} \\
\Rightarrow && y &= -px + ap^3+2ap \\
&& 2an &= -pan^2 + ap^3 + 2ap \\
\Rightarrow && 0 &= pan^2+2an-ap(2+p^2) \\
\Rightarrow && n &= p, -\left ( p + \frac2{p}\right) \\
\Rightarrow && n &= -\left ( p + \frac2{p}\right)
\end{align*}
\item $\,$ \begin{align*}
&& |PN|^2 &= (ap^2-an^2)^2 +(2ap-2an)^2 \\
&&&= a^2(p-n)^2(p+n)^2+4a^2(p-n)^2 \\
&&&= a^2(p-n)^2((p+n)^2+4) \\
&&&= a^2\left(p+p+\frac2p \right)^2 \left ( \left ( -\frac2p\right)^2+4\right)\\
&&&= a^2\left(\frac{2p^2+2}p \right)^2 \left ( \frac{4}{p^2}+4\right)\\
&&&= 16a^2 \frac{(p^2+1)^3}{p^4} \\
\\
&& \frac{\d |PN|^2}{\d p^2} &= 16a^2\frac{3(p^2+1)^2p^4-2(p^2+1)^3p^2}{p^8} \\
&&&= 16a^2(p^2+1)^2 \frac{3p^2-2(p^2+1)}{p^6} \\
&&&= 16a^2(p^2+1)^2 \frac{p^2-2}{p^6}
\end{align*}
Therefore minimized when $p^2=2$ (clearly a minimum by considering behaviour as $p^2 \to 0, \infty$)
\item If $PN$ is the diameter of $PNQ$ then $QP$ and $QN$ are perpendicular, ie
\begin{align*}
&& -1 &= \frac{2ap-2aq}{ap^2-aq^2} \cdot \frac{2aq-2an}{aq^2-an^2} \\
&&&= \frac{2}{p+q} \cdot \frac{2}{q+n} \\
&&&= \frac{2}{p+q} \cdot \frac{2}{q - p -\frac{2}{p}} \\
\Rightarrow && 4 &= (p+q)(p+\frac2{p}-q) \\
&&&= p^2-q^2 + \frac{2q}{p} + 2 \\
\Rightarrow && 2 &= p^2 - q^2 + \frac{2q}{p}
\end{align*}
Therefore $q = 0 \Rightarrow p^2 = 2 \Rightarrow |PN|$ is at it's minimum.
\end{questionparts}
Attempts at this question were over the thousand figure, making it the third most popular question on the paper, with the second‐highest mean score. Part (i) proved to be very routine; the calculus requirements in (ii) were obvious to most, though justifying the minimum distance was often poorly handled; for instance, finding the second derivative is a poor way to spend one's time when examining the sign of the first derivative is easily undertaken. The needs of part (iii) were also easily spotted though, again, a couple of marks were almost universally lost as the need to eliminate the two other cases that arise was largely ignored.