A2 June 2021 Paper 1 Q14
14 The hyperbola \(H\) has equation \(y^2 - x^2 = 16\)
The circle \(C\) has equation \(x^2 + y^2 = 32\)
The diagram below shows part of the graph of \(H\) and part of the graph of \(C\).
Show that the shaded region in the first quadrant enclosed by \(H\), \(C\), the \(x\)-axis and the \(y\)-axis has area
\[\frac{16\pi}{3} + 8\ln\left(\frac{\sqrt{2} + \sqrt{6}}{2}\right)\][12 marks]
| Scheme | Marks | AO |
|---|---|---|
| Find \(x\)-coordinate of P using simultaneous equations | M1 | 3.1a |
| Obtains correct \(x\)-coordinate of P | A1 | 1.1b |
| Splits region into two or more parts, at least one of which is given as an integral. All integrals with correct limits. Follow through their \(x\)-coordinate of P | M1 | 3.1a |
| Makes appropriate substitution to obtain A1 | M1 | 3.1a |
| Obtains correct integrand in terms of \(u\) Condone incorrect/omission of limits | A1 | 1.1b |
| Uses hyperbolic identity to integrate | M1 | 3.1a |
| Deduces that \(\sinh 2u = \sqrt{3}\) | M1 | 2.2a |
| Obtains correct value of A1 | A1 | 1.1b |
| Subtracts area of triangle from area of sector to obtain value of A2 or makes appropriate substitution to obtain A2 | M1 | 3.1a |
| Deduces that OP makes an angle of \(\dfrac{\pi}{3}\) with the \(x\)-axis or \(\left[\sin 2w\right]_{\pi/6}^{\pi/2} = \dfrac{-\sqrt{3}}{2}\) | M1 | 2.2a |
| Obtains correct value of A2 | A1 | 1.1b |
| Uses a rigorous argument by adding together the two areas | R1 | 2.1 |
| (12 marks) |
Typical solution
At P, \(y^2 - x^2 = 16\) and \(x^2 + y^2 = 32\)
\[x = 2\sqrt{2}\]\[A_1 = \int_0^{2\sqrt{2}} (x^2 + 16)^{\frac{1}{2}}\,\mathrm{d}x\]Let \(x = 4\sinh u\)
Then \((x^2 + 16)^{\frac{1}{2}} = 4\cosh u\)
and \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = 4\cosh u\)
\[A_1 = \int_0^{x = 2\sqrt{2}} 16\cosh^2 u\,\mathrm{d}u\]\[= 8\int_0^{x = 2\sqrt{2}} (\cosh 2u + 1)\,\mathrm{d}u\]\[= \left[4\sinh 2u + 8u\right]_0^{x = 2\sqrt{2}}\]When \(x = 2\sqrt{2}\),
\(\sinh u = \dfrac{\sqrt{2}}{2}\) and \(\cosh u = \dfrac{\sqrt{6}}{2}\)
\[\therefore \sinh 2u = 2\sinh u\cosh u = \sqrt{3}\]\[\text{So } A_1 = 4\sqrt{3} + 8\sinh^{-1}\left(\frac{\sqrt{2}}{2}\right)\]\[A_1 = 4\sqrt{3} + 8\ln\left(\frac{\sqrt{2} + \sqrt{6}}{2}\right)\]OP makes an angle of \(\dfrac{\pi}{3}\) with the \(x\)-axis
\[\text{So area of sector} = \frac{1}{2} \times 32 \times \frac{\pi}{3} = \frac{16\pi}{3}\]\[A_2 = \frac{16\pi}{3} - \frac{1}{2} \times 2\sqrt{2} \times 2\sqrt{6}\]\[A_2 = \frac{16\pi}{3} - 4\sqrt{3}\]\[\text{Required area} = A_1 + A_2\]\[= \frac{16\pi}{3} + 8\ln\left(\frac{\sqrt{2} + \sqrt{6}}{2}\right)\]as required