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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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