Year: 2022
Paper: 3
Question Number: 1
Course: LFM Pure and Mechanics
Section: Parametric equations
One question was attempted by well over 90% of the candidates two others by about 90%, and a fourth by over 80%. Two questions were attempted by about half the candidates and a further three questions by about a third of the candidates. Even the other three received attempts from a sixth of the candidates or more, meaning that even the least popular questions were markedly more popular than their counterparts in previous years. Nearly 90% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Let $C_1$ be the curve given by the parametric equations
\[ x = ct\,, \quad y = \frac{c}{t}\,, \]
where $c > 0$ and $t \neq 0$, and let $C_2$ be the circle
\[ (x-a)^2 + (y-b)^2 = r^2\,. \]
$C_1$ and $C_2$ intersect at the four points $P_i$ ($i = 1,2,3,4$), and the corresponding values of the parameter $t$ at these points are $t_i$.
\begin{questionparts}
\item Show that $t_i$ are the roots of the equation
\[ c^2 t^4 - 2act^3 + (a^2 + b^2 - r^2)t^2 - 2bct + c^2 = 0\,. \qquad (*) \]
\item Show that
\[ \sum_{i=1}^{4} t_i^2 = \frac{2}{c^2}(a^2 - b^2 + r^2) \]
and find a similar expression for $\displaystyle\sum_{i=1}^{4} \frac{1}{t_i^2}$.
\item Hence show that $\displaystyle\sum_{i=1}^{4} OP_i^2 = 4r^2$, where $OP_i$ denotes the distance of the point $P_i$ from the origin.
\item Suppose that the curves $C_1$ and $C_2$ touch at two distinct points.
By considering the product of the roots of $(*)$, or otherwise, show that the centre of circle $C_2$ must lie on either the line $y = x$ or $y = -x$.
\end{questionparts}\begin{questionparts}
\item Suppose $(ct, c/t)$ is on $C_2$ then
\begin{align*}
&& r^2 &= \left ( ct - a \right)^2 + \left ( \frac{c}{t} - b \right)^2 \\
&&&= c^2t^2 - 2cta + a^2 + \frac{c^2}{t^2} - \frac{2cb}{t} + b^2 \\
\Rightarrow && 0 &= c^2t^4 - 2act^3 + (a^2+b^2-r^2)t^2 - 2bct + c^2
\end{align*}
\item Notice that $\displaystyle \sum t_i = \frac{2a}{c}$ and $\displaystyle \sum t_it_j = \frac{a^2+b^2-r^2}{c^2}$ so
\begin{align*}
&& \sum t_i^2 &= \left ( \sum t_i \right)^2 - 2 \sum t_it_j \\
&&&= \frac{4a^2}{c^2} - \frac{2a^2+2b^2-2r^2}{c^2} \\
&&&= \frac{2}{c^2} \left (a^2 - b^2 + r^2 \right)
\end{align*}
Note that $\frac{1}{t}$ are roots of the $c^2 - 2act + (a^2+b^2-r^2)t^2 - 2bct^3 + c^2t^4 = 0$ which is the same equation but with $a \leftrightarrow b$ so $\displaystyle \sum \frac{1}{t_i^2} = \frac{2}{c^2} (b^2 - a^2 + r^2)$
\item Therefore
\begin{align*}
&& \sum_{i=1}^4 OP_i^2 &= \sum_{i=1}^4 \left (c^2t_i^2 + \frac{c^2}{t_i^2} \right) \\
&&&= 2(a^2-b^2+r^2) + 2(b^2-a^2+r^2) \\
&&&= 4r^2
\end{align*}
\item If they touch at two distinct points it must be the case that $t_1 = t_2$ and $t_3 = t_4$. We must also have $t_1t_2t_3t_4 = t_1^2t_3^2 = 1$ so $t_1t_3 = \pm 1$. Therefore our points are $(ct_1, \frac{c}{t_1})$ and $\pm(\frac{c}{t_1}, ct_1)$ but these are reflections in $y = \pm x$. But if these two points are reflections of one another the line of reflection is the perpendicular bisector, which must run through the centre of the circle.
\end{questionparts}
This was the most popular question with 94% attempting it, and it was also the most successful with a mean mark of nearly 14/20. Apart from very occasional inaccuracies, part (i) was always successfully done. The first summation result in part (ii) was usually successfully done, though there was some poor summation notation which let some candidates down. The second summation was completed successfully by virtue of some heavy algebra or, more efficiently, by seeing the connection to the first result dividing the quartic by t4 and comparing the quartic for the reciprocals with that in part (i). Some candidates were penalised for not justifying their result, having clearly worked backwards from part (iii). Part (iii) was well done except when candidates disregarded their result from part (ii). A lot of candidates managed to correctly interpret the implication of the curves touching at two distinct points in terms of the roots for t and the consequent result for the product of the four roots, but then struggled to reach the required result by algebra or poorly justified geometric arguments.