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2024 Paper 2 Q7
D: 1500.0
B: 1500.0
- Sketch the curve \(C_1\) with equation \[ \left(y^2 + (x-1)^2 - 1\right)\left(y^2 + (x+1)^2 - 1\right) = 0. \]
- Consider the curve \(C_2\) with equation
\[ \left(y^2 + (x-1)^2 - 1\right)\left(y^2 + (x+1)^2 - 1\right) = \tfrac{1}{16}. \]
- Show that the line \(y = k\) meets the curve \(C_2\) at points for which \[ x^4 + 2(k^2 - 2)x^2 + \left(k^4 - \tfrac{1}{16}\right) = 0. \] Hence determine the number of intersections between curve \(C_2\) and the line \(y = k\) for each positive value of \(k\).
- Determine whether the points on curve \(C_2\) with the greatest possible \(y\)-coordinate are further from, or closer to, the \(y\)-axis than those on curve \(C_1\).
- Show that it is not possible for both \(y^2 + (x-1)^2 - 1\) and \(y^2 + (x+1)^2 - 1\) to be negative, and deduce that curve \(C_2\) lies entirely outside curve \(C_1\).
- Sketch the curves \(C_1\) and \(C_2\) on the same axes.
Solution:
- \(\,\)
- Suppose \(y=k\) meets the curve \(C_2\) then \begin{align*} && \tfrac1{16} &= (k^2+(x-1)^2-1)(k^2+(x+1)^2-1) \\ &&&= (k^2+x^2-2x)(k^2+x^2+2x) \\ &&&= k^4+2k^2x^2+x^4-4x^2 \\ &&&= x^4+(2k^2-4)x^2+k^4 \\ \Rightarrow && 0 &= x^4+(2k^2-4)x^2+(k^4-\tfrac1{16}) \\ \\ && \Delta &= 4(k^2-2)^2 - 4 \cdot 1 \cdot (k^4-\tfrac1{16}) \\ &&&= 4(k^4-4k^2+4 - k^4 +\tfrac{1}{16}) \\ &&&= 16(1+\tfrac{1}{64} - k^2) \\ &&&= 16(\tfrac{65}{64} - k^2) \end{align*} Therefore if \(|k| < \frac{\sqrt{65}}{8}\) there are \(4\) intersections. If \(|k| = \frac{\sqrt{65}}{8}\) there are \(2\) intersections, otherwise there are \(0\).
- The greatest possible \(y\) value is \( \frac{\sqrt{65}}{8}\) and at this point \(x^2 = \frac{2(2-\frac{65}{64}}{2} =1 - \frac{1}{64} < 1\) so they are close to the \(y\)-axis.
- The regions where \(y^2+(x-1)^2-1 < 0\) and \(y^2+(x+1)^2-1 < 0\) is the interior of the two circles from the first part. However, since they don't overlap they can never both be negative. Therefore in our equation both are positive and therefore \(C_2\) is entirely outside \(C_1\)
- \(\,\)
