Year: 2016
Paper: 3
Question Number: 2
Course: UFM Pure
Section: Conic sections
A substantially larger number of candidates took the paper this year: 14% more than in 2015. However, the mean score was virtually identical to that in 2015. Five questions were very popular, with two being attempted by in excess of 90% of the candidates, but once again, all questions were attempted by significant numbers, with only one dipping under 10% attempting it, and every question was answered perfectly by at least one candidate. Most candidates kept to six sensible attempts, although some did several more scoring weakly overall, except in six outstanding cases that earned very high marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
The distinct points $P(ap^2 , 2ap)$, $Q(aq^2 , 2aq)$ and $R(ar^2,2ar)$ lie on the parabola $y^2 = 4ax$, where $a>0$. The points are such that the normal to the parabola at $Q$ and the normal to the parabola at $R$ both pass through $P$.
\begin{questionparts}
\item Show that $q^2 +qp + 2 = 0$.
\item
Show that $QR$ passes through a certain point that is independent of the choice of $P$.
\item
Let $T$ be the point of intersection of $OP$ and $QR$, where $O$ is the coordinate origin. Show that $T$ lies on a line that is independent of the choice of $P$.
Show further that the distance from the $x$-axis to $T$ is less than $\dfrac {\;a}{\sqrt2}\,$.
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& 2y \frac{\d y}{\d x} &= 4a \\
\Rightarrow && \frac{\d y}{\d x} &= \frac{2a}{y}
\end{align*}
Therefore we must have
\begin{align*}
&& \underbrace{-\frac{2aq}{2a}}_{\text{gradient of normal}} &= \underbrace{\frac{2ap-2aq}{ap^2-aq^2}}_{\Delta y / \Delta x} \\
\Rightarrow && -q &= \frac{2}{p+q} \\
&& 0 &= 2 + pq+q^2
\end{align*}
\item We must have that $q,r$ are the two roots of $x^2+px+2 = 0$
$QR$ has the equation:
\begin{align*}
&& \frac{y-2aq}{x-aq^2} &= \frac{2ar-2aq}{ar^2-aq^2} \\
\Rightarrow && \frac{y-2aq}{x-aq^2} &= \frac{2}{r+q} \\
\Rightarrow && y &= \frac{2}{q+r}(x-aq^2) +2aq \\
&& y &= -\frac{2}{p}x+2a\left(q-\frac{q^2}{q+r} \right) \\
&&y&= -\frac{2}{p}x+2a \frac{qr}{q+r} \\
&& y &= -\frac{2}{p}x - 2a \frac{2}{p} \\
&& y & = -\frac{2}{p}(x+2a)
\end{align*}
Therefore the point $(-2a,0)$ lies on all such lines.
\item
$OP$ has equation $y = \frac{2}{p} x$
\begin{align*}
&& y &= \frac{2}{p} x \\
&& y & = -\frac{2}{p}(x+2a) \\
&& 2y &= -\frac{4a}{p} \\
\Rightarrow && y &= -\frac{2a}{p} \\
&& x &= -a
\end{align*}
Therefore $T\left (-a, -\frac{2a}{p} \right)$ always lies on the line $x = -a$
The distance to the $x$-axis from $T$ is $\frac{2a}{|p|}$. We need to show that $p$ can't be too small. Specifically $x^2+px+2 = 0$ must have $2$ real roots, ie $\Delta = p^2-8 \geq 0 \Rightarrow |p| \geq 2\sqrt{2}$, ie $\frac{2a}{|p|} \leq \frac{2a}{2\sqrt{2}} = \frac{a}{\sqrt{2}}$ as required.
\end{questionparts}
This question was quite popular, being attempted by just over three-quarters of the candidates, but success was moderate. Most got underway differentiating implicitly, rather than parametrically, and were able to find equations of tangents, normal and chords, but not always simplifying by factorising to make their lives easier; those who could factorise made good progress, whilst those who did not struggled to find (i). Other weaknesses were not appreciating that they could find r+q and rq, which then led to not finding the certain point in (ii). In the final part, square rooting the inequality and only considering the positive case was not uncommon.