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2022 Paper 3 Q1
D: 1500.0 B: 1500.0
parametric equations curve intersection vieta's formulas roots of polynomials circle rectangular hyperbola symmetric functions of roots

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 \(t_i\).

  1. Show that \(t_i\) are the roots of the equation \[ c^2 t^4 - 2act^3 + (a^2 + b^2 - r^2)t^2 - 2bct + c^2 = 0\,. \qquad (*) \]
  2. 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 \(\displaystyle\sum_{i=1}^{4} \frac{1}{t_i^2}\).
  3. 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.
  4. 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 \(y = -x\).

Solution:

  1. 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*}
  2. 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 \(\displaystyle \sum \frac{1}{t_i^2} = \frac{2}{c^2} (b^2 - a^2 + r^2)\)
  3. 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*}
  4. 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.

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