Year: 2012
Paper: 1
Question Number: 4
Course: LFM Pure and Mechanics
Section: Parametric equations
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1500.0
Banger Comparisons: 0
The curve $C$ has equation $xy = \frac12$.
The tangents to $C$ at the distinct points $P\big(p, \frac1{2p}\big)$ and $Q\big(q, \frac1{2q}\big)$ where $p$ and $q$ are positive, intersect at $T$ and the normals to $C$ at these points intersect at $N$. Show that $T$ is the point
\[
\left( \frac{2pq}{p+q}\,,\, \frac 1 {p+q}\right)\!.
\]
In the case $pq=\frac12$, find the coordinates of $N$. Show (in this case) that $T$ and $N$ lie on the line $y=x$ and are such that the
product of their distances from the origin is constant.\begin{align*}
&& \frac{\d y}{\d x} &= -\frac1{2x^2} \\
\Rightarrow && \frac{y - \frac{1}{2p}}{x - p} &= - \frac{1}{2p^2} \\
\Rightarrow && y - \frac1{2p} &= -\frac{1}{2p^2}x +\frac1{2p} \\
\Rightarrow && y +\frac{1}{2p^2}x &= \frac1p \\
\Rightarrow && 2p^2 y + x &= 2p\\
\Rightarrow && 2q^2 y + x &= 2q \\
\Rightarrow && (p^2-q^2)y &= p-q \\
\Rightarrow && y &= \frac{1}{p+q} \\
&& x &= \frac{2pq}{p+q}
\end{align*}
\begin{align*}
\text{normal}: && \frac{y-\frac1{2p}}{x-p} &= 2p^2 \\
\Rightarrow && y - \frac1{2p} &= 2p^2x - 2p^3 \\
\Rightarrow && 2py -4p^3x &= 1-4p^4 \\
\Rightarrow && 2qy -4q^3x &= 1-4q^4 \\
pq = \tfrac12: && y - 2p^2 x &= q-2p^3 \\
&& y - 2q^2 x &= p-2q^3 \\
\Rightarrow && (2q^2-2p^2)x &= q-p +2q^3-2p^3 \\
&&&= (q-p)(q+p+2q^2+1+2p^2) \\
\Rightarrow && x &= \frac{1+2(p^2+q^2)+1}{2(p+q)} \\
&&&= \frac{1+2(p^2+q^2+2pq-1)+1}{2(p+q)} \\
&&&= p+q\\
&& y &= 2p^2 \left ( p+q \right) + q - 2p^3 \\
&&&= p+q
\end{align*}
So $N(p+q, p+q)$ and $T\left (\frac{1}{p+q}, \frac{1}{p+q} \right)$, so both points lie on $y = x$.
\[ OT \cdot ON = \frac{\sqrt{2}}{p+q} \cdot (p+q)\sqrt{2} = 2 \]
which is clearly constant.
This question was generally well attempted, with many candidates able to obtain the equations of the tangents and normals. The point of intersection of the tangents did not cause too much difficulty, but the intersection of the normal was problematic for many candidates who struggled to simplify the expressions and therefore did not reach a point where it was obvious that the point lay on the line. The final part of the question is much easier if the expressions are simplified as they are encountered and for this the factorisation of the difference of two cubes is a useful thing to know. It is also useful to realise that the information that means that the expression can be factorised as.