A2 June 2019 Paper 2 Q8
8 A parabola \(P_1\) has equation \(y^2 = 4ax\) where \(a \gt 0\)
\(P_1\) is translated by the vector \(\begin{bmatrix} b \\ 0 \end{bmatrix}\), where \(b \gt 0\), to give the parabola \(P_2\)
Prove that \(m = \pm\sqrt{\dfrac{a}{b}}\)
Solutions using differentiation will be given no marks. [4 marks]
The finite region \(R\) is bounded by the \(x\)-axis, \(P_2\) and a line through \(D\) perpendicular to the \(x\)-axis.
The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis to form a solid.
Find, in terms of \(a\) and \(b\), the volume of this solid.
Fully justify your answer. [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains the equation of \(P_2\) | B1 | 1.2 |
| Combines equations to give a quadratic equation in \(x\) | M1 | 1.1a |
| Sets the discriminant to zero, with working | M1 | 1.1a |
| Completes a rigorous argument to show the required result. | R1 | 2.1 |
Typical solution
\[y^2 = 4a(x - b)\]\[m^2x^2 = 4a(x - b)\]\[m^2x^2 - 4ax + 4ab = 0\]For equal roots \(\Delta = 0\)
\[16a^2 - 4m^2(4ab) = 0\]\[16a^2 = 16m^2ab\]\[m = \pm\sqrt{\frac{a}{b}}\]| Scheme | Marks | AO |
|---|---|---|
| Forms a quadratic equation in \(x\) | M1 | 3.1a |
| Finds correct \(x\) value at D | A1 | 1.1b |
| Writes a correct integral for V. Condone limits omitted | B1 | 1.2 |
| Correctly integrates their two-term expression with lower limit \(b\) | M1 | 1.1a |
| Finds the correct answer | A1 | 1.1b |
| (9 marks) |
Typical solution
\[\left(x\sqrt{\frac{a}{b}}\right)^2 = 4a(x - b)\]\[x^2a = 4ab(x - b)\]\[x^2 - 4bx + 4b^2 = 0\]\[(x - 2b)^2 = 0 \Longrightarrow x = 2b\]
Notes
[This can also be done by translating the curve by \(\begin{pmatrix} -b \\ 0 \end{pmatrix}\) and finding \(\pi\int_0^b 4ax\,\mathrm{d}x\), but this must be clearly explained to gain full marks.]