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2012 Paper 3 Q3
It is given that the two curves \[ y=4-x^2 \text{ and } m x = k-y^2\,, \] where \(m > 0\), touch exactly once.
- In each of the following four cases, sketch the two curves on a single diagram, noting the coordinates of any intersections with the axes:
- \(k < 0\, \);
- \(0 < k < 16\), \(k/m < 2\,\);
- \(k > 16\), \(k/m > 2\,\);
- \(k > 16\), \(k/m < 2\,\).
- Now set \(m=12\). Show that the \(x\)-coordinate of any point at which the two curves meet satisfies \[ x^4-8x^2 +12x +16-k=0\,. \] Let \(a\) be the value of \(x\) at the point where the curves touch. Show that \(a\) satisfies \[ a^3 -4a +3 =0 \] and hence find the three possible values of \(a\). Derive also the equation \[ k= -4a^2 +9a +16\,. \] Which of the four sketches in part (i) arise?
Solution:
- \(\,\)
- \(\,\)
- \(\,\)
- \(\,\)
- \(\,\)
- Suppose \(m = 12\) \begin{align*} && y &= 4-x^2 \\ && 12x &= k-y^2 \\ \Rightarrow && 12 x&=k-(4-x^2)^2 \\ &&&= k-16+8x^2-x^4 \\ \Rightarrow && 0 &= x^4- 8x^2+12x+16-k \end{align*} When the curves touch, we will have repeated root, ie \(a\) is a root of \(4x^3-16x+12 \Rightarrow a^3-4a+3 =0\). \begin{align*} &&0 &= a^3-4a+3 \\ &&&= (a-1)(a^2+a-3) \\ \Rightarrow &&a &= 1, \frac{-1 \pm \sqrt{13}}{2} \end{align*} \begin{align*} && 0 &= a^4-8a^2+12a+16-k \\ \Rightarrow && k &= a(a^3-8a+12)+16 \\ &&&= a(4a-3-8a+12)+16 \\ &&&= -4a^2+9a+16 \\ \\ \Rightarrow && a = 1& \quad k = 21 \\ && k &= -4(3-a)+9a+16 = 13a+4\\ && a = \frac{-1-\sqrt{13}}2& \quad k = \frac{-5 - 13\sqrt{13}}{2} < 0 \\ && a = \frac{-1+\sqrt{13}}2& \quad k = \frac{-5 + 13\sqrt{13}}{2} \\ \end{align*} So we have type (a), and (d).
2007 Paper 1 Q8
A curve is given by the equation \[ y = ax^3 - 6ax^2+ \left( 12a + 12 \right)x - \left( 8a + 16 \right)\,, \tag{\(*\)} \] where \(a\) is a real number. Show that this curve touches the curve with equation \[ y=x^3 \tag{\(**\)} \] at \(\left( 2 \, , \, 8 \right)\). Determine the coordinates of any other point of intersection of the two curves.
- Sketch on the same axes the curves \((*)\) and \((**)\) when \(a = 2\).
- Sketch on the same axes the curves \((*)\) and \((**)\) when \(a = 1\).
- Sketch on the same axes the curves \((*)\) and \((**)\) when \(a = -2\).
Solution: \begin{align*} && y &= ax^3 - 6ax^2+ \left( 12a + 12 \right)x - \left( 8a + 16 \right) \\ && y(2) &= 8a-24a+24a+24-8a-16 \\ &&&= 8 \\ && y'(x) &= 3ax^2-12ax+(12a+12) \\ && y'(0) &= 12a-24a+12a+12 \\ &&&= 12 \end{align*} Therefore since our curve has the same value and gradient at \((2,8)\) as \(y = x^3\) they must touch at this point. Therefore \begin{align*} && ax^3 - 6ax^2+ \left( 12a + 12 \right)x - \left( 8a + 16 \right) - x^3 &= (x-2)^2((a-1)x-(2a+4)) \end{align*} Therefore if \(a \neq 1\), they touch again when \(x = \frac{2a+4}{a-1}\).
