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2013 Paper 1 Q5
D: 1500.0 B: 1470.2
conic sections locus curve sketching rotation of axes factorisation circle parabola coordinate geometry transformations

The point \(P\) has coordinates \((x,y)\) which satisfy \[ x^2+y^2 + kxy +3x +y =0\,. \]

  1. Sketch the locus of \(P\) in the case \(k=0\), giving the points of intersection with the coordinate axes.
  2. By factorising \(3x^2 +3y^2 +10xy\), or otherwise, sketch the locus of \(P\) in the case \(k=\frac{10}{3}\,\), giving the points of intersection with the coordinate axes.
  3. In the case \(k=2\), let \(Q\) be the point obtained by rotating \(P\) clockwise about the origin by an angle~\(\theta\), so that the coordinates \((X,Y)\) of \(Q\) are given by \[ X=x\cos\theta +y\sin\theta\,, \ \ \ \ Y= -x\sin\theta + y\cos\theta\,. \] Show that, for \(\theta =45^\circ\), the locus of \(Q\) is \( \sqrt2 Y= (\sqrt2 X+1 )^2 - 1 .\) Hence, or otherwise, sketch the locus of \(P\) in the case \(k=2\), giving the equation of the line of symmetry.

Solution:

  1. \(k = 0\), we have \(x^2 + y^2 + 3x + y = 0\), ie \((x+\tfrac32)^2+(y+\tfrac12)^2 = \frac{10}{4}\).
    TikZ diagram
  2. \(3x^2 + 3y^2 +10xy = (3x+y)(x+3y)\) so \(x^2 + y^2 + \tfrac{10}3xy + 3x+y = (3x+y)(\frac{x+3y}{3}+1) = 0\) so we have the line pair \(3x +y =0\), \(x+3y + 3 = 0\)
    TikZ diagram
  3. If \(k = 2\) then \((x+y)^2 + (x+y)+2x = 0\). If \(\theta = 45^\circ\) then \( X = \frac1{\sqrt{2}}(x+y), Y = \frac{1}{\sqrt{2}}(y-x)\), ie \(x+y = \sqrt{2}X\) and \(x = \frac{1}{\sqrt2}(X-Y)\), so our equation is: \begin{align*} 0 &= 2X^2 + \sqrt{2}X + \sqrt{2}(X-Y) \\ &= (\sqrt{2}X + 1)^2 - 1 - \sqrt{2} Y \end{align*} which would be a parabola with line of symmetry \(X = -\frac{1}{\sqrt{2}}\). However, we are actually looking at that parabola rotated by \(45^\circ\) anticlockwise.
    TikZ diagram

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1989 Paper 2 Q8
D: 1665.0 B: 1499.2
groups complex numbers transformations permutations roots of unity group order Möbius transformation symmetry group

Let \(\Omega=\exp(\mathrm{i}\pi/3).\) Prove that \(\Omega^{2}-\Omega+1=0.\) Two transformations, \(R\) and \(T\), of the complex plane are defined by \[ R:z\longmapsto\Omega^{2}z\qquad\mbox{ and }\qquad T:z\longmapsto\dfrac{\Omega z+\Omega^{2}}{2\Omega^{2}z+1}. \] Verify that each of \(R\) and \(T\) permute the four point \(z_{0}=0,\) \(z_{1}=1,\) \(z_{2}=\Omega^{2}\) and \(z_{3}=-\Omega.\) Explain, without explicitly producing a group multiplication table, why the smallest group of transformations which contains elements \(R\) and \(T\) has order at least 12. Are there any permutations of these points which cannot be produced by repeated combinations of \(R\) and \(T\)?

Solution: \(R(0) = 0\), \(R(1) = \Omega^2 1 = \Omega^2\), \(R(\Omega^2) = \Omega^4 = -\Omega\), \(R(-\Omega) = -\Omega^3 = 1\) \(T(0) = \frac{\Omega^2}1 = \Omega^2\), \(T(1) = \frac{\Omega + \Omega^2}{2\Omega^2+1} = \frac{2\Omega - 1}{2\Omega-1} = 1\) \(T(\Omega^2) = \frac{\Omega^3 + \Omega^2}{2\Omega^4+1} = \Omega \frac{\Omega^2+\Omega}{-2\Omega+1} = \Omega \frac{2\Omega-1}{-2\Omega+1} = - \Omega\) \(T(-\Omega) = \frac{-\Omega^2 + \Omega^2}{-2\Omega^3+1} = \frac{0}{3} = 0\) Thinking of \(R\) and \(S\) as elements of \(S_4\), we have that \(R = (234), S = (134)\), we can also construct \(RS = (14)(23), R^2S = (12)(34), RSR^2S = (13)(24)\). Therefore we have the subgroups \(\{e, (234), (243)\}\) of order \(3\) and the subgroup \(\{e, (12)(34), (13)(24), (14)(23) \}\) of order \(4\). By Lagrange's theorem this means that both \(3\) and \(4\) divide the order of the group, therefore the group has order divisible by \(12\) (and therefore is at least \(12\)). Yes, we cannot produce any odd permutation, for example \((12)\) cannot be produced. (Since all our generators are even permutations).

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