Challenging +1.8 This is a Further Maths question requiring integration to find an area between curves. Students must find intersection points, set up integrals for both the hyperbola and circle portions, evaluate them (including inverse hyperbolic functions or substitution for the hyperbola), and combine results. While technically demanding with multiple steps and requiring careful handling of the hyperbola integral, it follows a standard 'show that' format with a clear roadmap and uses techniques expected at this level.
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 \(C\).
\includegraphics{figure_14}
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]
Question 14:
14 | Find x-coordinate of P using
simultaneous equations | 3.1a | M1 | At P, and
2 2 2 2
π¦π¦ βπ₯π₯ = 16 π₯π₯ +π¦π¦ = 32
π₯π₯ = 2β2
2β2 1
2 2
Let π΄π΄1 = οΏ½ (π₯π₯ +16) dπ₯π₯
0
Then
π₯π₯ = 4sinhπ’π’1
and 2 2
(π₯π₯ +16) = 4coshπ’π’
dπ₯π₯
dπ’π’ = 4coshπ’π’
π₯π₯=2β2
2
π΄π΄1 = οΏ½ 16cosh π’π’dπ’π’
0
π₯π₯=2β2
= 8οΏ½ (cosh2π’π’+1)d π’π’
0
When , π₯π₯=2β2
= [4sinh2π’π’+8π’π’]0
and
π₯π₯ = 2β2
β2 β6
sinhπ’π’ = 2 coshπ’π’ = 2
So
β΄ sinh2π’π’ = 2sinhπ’π’coshπ’π’ = β3
β1 β2
π΄π΄1 = 4β3+8sinh οΏ½2οΏ½
β2+β6
π΄π΄1 = 4β3+8lnοΏ½ οΏ½
2
OP makes an angle of with the x-axis
ππ
So area of sector
3
1 ππ 16ππ
= 2Γ32Γ3 = 3
16ππ 1
π΄π΄2 = β Γ2β2Γ 2β6
3 2
Required area 16ππ
π΄π΄2 = β4β3
3
= π΄π΄1+π΄π΄2
16ππ β2+β6
as require=d +8lnοΏ½ οΏ½
3 2
Obtains correct x-coordinate
of P | 1.1b | A1
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 | 3.1a | M1
Makes appropriate
substitution to obtain A
1 | 3.1a | M1
Obtains correct integrand in
terms of u
Condone incorrect/omission
of limits | 1.1b | A1
Uses hyperbolic identity to
integrate | 3.1a | M1
Deduces that | 2.2a | M1
sinh2π’π’ = β3
Obtains correct value of A
1 | 1.1b | A1
Subtracts area of triangle
from area of sector to obtain
value of A
2
or
makes appropriate
substitution to obtain A
2 | 3.1a | M1
Deduces that
OP makes an angle of with
ππ
the x-axis
3
or | 2.2a | M1
ππβ2 ββ3
[sin2π€π€]ππβ6 =
Obtains correct value2 of A
2 | 1.1b | A1
Uses a rigorous argument by
adding together the two
areas | 2.1 | R1
Total | 12
Q | Marking Instructions | AO | Marks | Typical solution
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 $C$.
\includegraphics{figure_14}
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]
\hfill \mbox{\textit{AQA Further Paper 1 2021 Q14 [12]}}