Year: 2018
Paper: 3
Question Number: 4
Course: UFM Pure
Section: Polar coordinates
The total entry was a record number, an increase of over 6% on 2017. Only question 1 was attempted by more than 90%, although question 2 was attempted by very nearly 90%. Every question was attempted by a significant number of candidates with even the two least popular questions being attempted by 9%. More than 92% restricted themselves to attempting no more than 7 questions with very few indeed attempting more than 8. As has been normal in the past, apart from a handful of very strong candidates, those attempting six questions scored better than those attempting more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1503.2
Banger Comparisons: 2
The point $P(a\sec \theta, b\tan \theta )$ lies on the hyperbola
\[
\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\,,
\]
where $a>b>0\,$.
Show that the equation of the tangent to the hyperbola at $P$ can be written as
\[
bx- ay \sin\theta = ab \cos\theta
\,.
\]
\begin{questionparts}
\item This tangent meets the
lines
$\dfrac x a = \dfrac yb$ and $\dfrac x a =- \dfrac y b$
at $S$ and $T$, respectively.
How is the mid-point of $ST$ related to $P$?
\item
The point $Q(a\sec \phi, b\tan \phi)$ also lies on the hyperbola and the tangents to the hyperbola at $P$ and $Q$ are perpendicular.
These two tangents intersect at $(x,y)$.
Obtain expressions for $x^2$ and $y^2$ in terms of $a$, $\theta$ and $\phi$.
Hence, or otherwise, show that
$x^2+y^2 = a^2 -b^2$.
\end{questionparts}Note that
\begin{align*}
&& \frac{\d a \sec \theta}{\d \theta} &= a \sec \theta \tan \theta \\
&& \frac{\d b \tan \theta}{\d \theta} &= b \sec^2 \theta \\
\Rightarrow && \frac{\d y}{\d x} &= \frac{b \sec^2 \theta}{a \sec \theta \tan \theta} \\
&&&= \frac{b}{a} \frac{1}{\sin \theta} \\
\Rightarrow && \frac{y-b \tan \theta}{x - a \sec \theta} &= \frac{b}{a} \frac{1}{\sin \theta} \\
\Rightarrow && a \sin \theta y - ab \tan \theta \sin \theta &= bx -ab \sec \theta \\
\Rightarrow && bx-ay\sin \theta &= ab \sec x (1 - \sin ^2 \theta) \\
&&&= ab \cos \theta
\end{align*}
\begin{questionparts}
\item \begin{align*}
S: &&& \begin{cases} bx-ay &= 0 \\
bx-ay \sin \theta &= ab \cos \theta \end{cases} \\
\Rightarrow && ay(1-\sin \theta) &= ab \cos \theta \\
\Rightarrow && y &= \frac{b \cos \theta}{1-\sin \theta} \\
&&x &=\frac{a\cos \theta}{1-\sin \theta} \\
T: &&& \begin{cases} bx+ay &= 0 \\
bx-ay \sin \theta &= ab \cos \theta \end{cases} \\
\Rightarrow && ay(1+\sin \theta) &= -ab \cos \theta \\
\Rightarrow && y &= \frac{-b \cos \theta}{1+\sin \theta} \\
&&x &=\frac{a\cos \theta}{1+\sin \theta} \\
M: && x &= \frac{a \cos \theta}{2} \frac{2}{1-\sin^2 \theta} \\
&&&= a \sec \theta \\
&& y &= \frac{b \cos \theta}{2} \frac{2 \sin \theta}{1-\sin^2 \theta} \\
&&&= b \tan \theta
\end{align*}
The midpoint of $ST$ is the same as $P$.
\item The tangents are perpendicular, therefore $\frac{b}{a} \cosec \theta = - \frac{a}{b} \sin \phi$, ie $b^2 = -a^2 \sin \phi \sin \theta$
The will intersect at:
\begin{align*}
&&& \begin{cases} bx - ay \sin \theta &= ab \cos \theta \\
bx - ay \sin \phi &= ab \cos \phi \end{cases} \\
\Rightarrow && ay ( \sin \theta - \sin \phi) &= ab(\cos \phi - \cos \theta) \\
\Rightarrow && y &= \frac{b(\cos \phi - \cos \theta)}{(\sin \theta - \sin \phi)} \\
&& y^2 &= \frac{-a^2 \sin \phi \sin \theta (\cos\phi - \cos \theta)^2}{(\sin \theta - \sin \phi)^2} \\
\Rightarrow && bx(\sin \phi - \sin \theta) &= ab(\cos \theta \sin \phi - \cos \phi \sin \theta) \\
\Rightarrow && x &= \frac{a(\cos \theta \sin \phi - \cos \phi \sin \theta)}{\sin \phi - \sin \theta} \\
&&&= \frac{a^2(\cos \theta \sin \phi - \cos \phi \sin \theta)^2}{(\sin \phi - \sin \theta)^2}
\end{align*}
Therefore
\begin{align*}
&& x^2+y^2 &= \frac{a^2}{(\sin \phi - \sin \theta)^2} \l (\cos \theta \sin \phi- \cos \phi \sin \theta)^2 - \sin \phi \sin \theta (\cos\phi - \cos \theta)^2 \r \\
&&&= \frac{a^2}{(\sin \phi - \sin \theta)^2} \l (\sin \phi - \sin \theta)(\cos^2 \theta \sin \phi - \sin \theta \cos^2 \phi) \r \\
&&&=a^2-b^2
\end{align*}
\end{questionparts}
The third most popular question being attempted by just short of three quarters of the candidature, it was however the most successfully attempted with a mean score of not quite 12/20. The stem was usually correctly attempted either using parametric or implicit differentiation. Simultaneous equations were sensibly attempted for part (i), but sometimes they confused the two pairs of equations and as a result got the wrong answer. Some solutions elegantly achieved the correct result having found just one of the coordinates and arguing that as it lay on the tangent, it had to be the point P. Part (ii) was quite often abandoned partway through, giving up after obtaining and in the face of the algebra, although some forgot to answer the question at this point even though they had employed simultaneous equations to obtain and. Few managed to conclude the question, and it was very rare indeed that the non-zero nature of the denominator sin sin was justified.