Question 8 - A-Level Maths

AQA Further Paper 2 2019 June — Question 8 9 marks

Exam Board	AQA
Module	Further Paper 2 (Further Paper 2)
Year	2019
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Conic sections
Type	Conic translation and transformation
Difficulty	Challenging +1.8 Part (a) requires algebraic manipulation with the discriminant condition for tangency (no differentiation allowed), which is non-standard for A-level. Part (b) involves volume of revolution with a parabolic boundary requiring careful setup and integration by substitution. The combination of geometric insight, algebraic proof constraints, and multi-step integration with parameters makes this substantially harder than typical Further Maths questions, though the individual techniques are accessible.
Spec	1.08d Evaluate definite integrals: between limits 4.08d Volumes of revolution: about x and y axes

A parabola \(P_1\) has equation \(y^2 = 4ax\) where \(a > 0\) \(P_1\) is translated by the vector \(\begin{bmatrix} b \\ 0 \end{bmatrix}\), where \(b > 0\), to give the parabola \(P_2\)

The line \(y = mx\) is a tangent to \(P_2\) Prove that \(m = \pm\sqrt{\frac{a}{b}}\) Solutions using differentiation will be given no marks. [4 marks]
The line \(y = \sqrt{\frac{a}{b}} x\) meets \(P_2\) at the point \(D\). 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]

Show mark scheme Show mark scheme source

Question 8:

Answer	Marks	Guidance
8(a)	Obtains the equation of P
2	AO1.2	B1

𝑦𝑦 = 4𝑎𝑎(𝑎𝑎−𝑏𝑏)

2 2

𝑚𝑚 𝑎𝑎 = 4𝑎𝑎(𝑎𝑎−𝑏𝑏)

For equal 2roo2ts

𝑚𝑚 𝑎𝑎 −4𝑎𝑎𝑎𝑎 +4𝑎𝑎𝑏𝑏 = 0

Δ = 0

2 2

16𝑎𝑎 −4𝑚𝑚 (4𝑎𝑎𝑏𝑏) = 0

2 2

16𝑎𝑎 = 16𝑚𝑚 𝑎𝑎𝑏𝑏

𝑎𝑎

𝑚𝑚 = ±�

𝑏𝑏

Combines equations to

give a quadratic equation

Answer	Marks	Guidance
in	AO1.1a	M1

Se𝑎𝑎ts the discriminant to

Answer	Marks	Guidance
zero, with working	AO1.1a	M1

Completes a rigorous

argument to show the

Answer	Marks	Guidance
required result.	AO2.1	R1
(b)	Forms a quadratic
equation in	AO3.1a	M1

𝑎𝑎

�𝑎𝑎� � = 4𝑎𝑎(𝑎𝑎−𝑏𝑏 )

𝑏𝑏

2 𝑎𝑎 𝑎𝑎 = 4𝑎𝑎𝑏𝑏(𝑎𝑎− 𝑏𝑏)

2 2

𝑎𝑎 −4 𝑏𝑏𝑎𝑎+4𝑏𝑏 = 0

2 (𝑎𝑎−2𝑏𝑏) = 0 ⟹ 𝑎𝑎 = 2𝑏𝑏

2𝑏𝑏

𝑉𝑉 = 𝜋𝜋� 4𝑎𝑎(𝑎𝑎−𝑏𝑏)𝑑𝑑𝑎𝑎

𝑏𝑏

2 2𝑏𝑏

𝑎𝑎

𝑉𝑉 = 4𝑎𝑎𝜋𝜋� −𝑏𝑏𝑎𝑎�

2 𝑏𝑏

2 2

4𝑏𝑏 2 𝑏𝑏 2

= 4𝑎𝑎𝜋𝜋�� −2𝑏𝑏 �− � −𝑏𝑏 ��

2 2

2 = 2𝑎𝑎𝑏𝑏 𝜋𝜋

Answer	Marks	Guidance
Finds correc𝑎𝑎t value at D	AO1.1b	A1

Writes a corre𝑎𝑎ct integral

for V. Condone limits

Answer	Marks	Guidance
omitted	AO1.2	B1

Correctly integrates their

two-term expression with

Answer	Marks	Guidance
lower limit b	AO1.1a	M1

Finds the correct answer

[This can also be done by

translating the curve by

and finding

−𝑏𝑏

� � , but this must

0 be c𝑏𝑏learly explained to

Answer	Marks	Guidance
g𝜋𝜋a∫i0n 4fu𝑎𝑎ll𝑎𝑎 m𝑑𝑑a𝑎𝑎rks.]	AO1.1b	A1
Total	9
Q	Marking Instructions	AO

Question 8:
--- 8(a) ---
8(a) | Obtains the equation of P
2 | AO1.2 | B1 | 2
𝑦𝑦 = 4𝑎𝑎(𝑎𝑎−𝑏𝑏)
2 2
𝑚𝑚 𝑎𝑎 = 4𝑎𝑎(𝑎𝑎−𝑏𝑏)
For equal 2roo2ts
𝑚𝑚 𝑎𝑎 −4𝑎𝑎𝑎𝑎 +4𝑎𝑎𝑏𝑏 = 0
Δ = 0
2 2
16𝑎𝑎 −4𝑚𝑚 (4𝑎𝑎𝑏𝑏) = 0
2 2
16𝑎𝑎 = 16𝑚𝑚 𝑎𝑎𝑏𝑏
𝑎𝑎
𝑚𝑚 = ±�
𝑏𝑏
Combines equations to
give a quadratic equation
in | AO1.1a | M1
Se𝑎𝑎ts the discriminant to
zero, with working | AO1.1a | M1
Completes a rigorous
argument to show the
required result. | AO2.1 | R1
(b) | Forms a quadratic
equation in | AO3.1a | M1 | 2
𝑎𝑎
�𝑎𝑎� � = 4𝑎𝑎(𝑎𝑎−𝑏𝑏 )
𝑏𝑏
2
𝑎𝑎 𝑎𝑎 = 4𝑎𝑎𝑏𝑏(𝑎𝑎− 𝑏𝑏)
2 2
𝑎𝑎 −4 𝑏𝑏𝑎𝑎+4𝑏𝑏 = 0
2
(𝑎𝑎−2𝑏𝑏) = 0 ⟹ 𝑎𝑎 = 2𝑏𝑏
2𝑏𝑏
𝑉𝑉 = 𝜋𝜋� 4𝑎𝑎(𝑎𝑎−𝑏𝑏)𝑑𝑑𝑎𝑎
𝑏𝑏
2 2𝑏𝑏
𝑎𝑎
𝑉𝑉 = 4𝑎𝑎𝜋𝜋� −𝑏𝑏𝑎𝑎�
2 𝑏𝑏
2 2
4𝑏𝑏 2 𝑏𝑏 2
= 4𝑎𝑎𝜋𝜋�� −2𝑏𝑏 �− � −𝑏𝑏 ��
2 2
2
= 2𝑎𝑎𝑏𝑏 𝜋𝜋
Finds correc𝑎𝑎t value at D | AO1.1b | A1
Writes a corre𝑎𝑎ct integral
for V. Condone limits
omitted | AO1.2 | B1
Correctly integrates their
two-term expression with
lower limit b | AO1.1a | M1
Finds the correct answer
[This can also be done by
translating the curve by
and finding
−𝑏𝑏
� � , but this must
0
be c𝑏𝑏learly explained to
g𝜋𝜋a∫i0n 4fu𝑎𝑎ll𝑎𝑎 m𝑑𝑑a𝑎𝑎rks.] | AO1.1b | A1
Total | 9
Q | Marking Instructions | AO | Marks | Typical solution

Show LaTeX source

A parabola $P_1$ has equation $y^2 = 4ax$ where $a > 0$

$P_1$ is translated by the vector $\begin{bmatrix} b \\ 0 \end{bmatrix}$, where $b > 0$, to give the parabola $P_2$

\begin{enumerate}[label=(\alph*)]
\item The line $y = mx$ is a tangent to $P_2$

Prove that $m = \pm\sqrt{\frac{a}{b}}$

Solutions using differentiation will be given no marks.
[4 marks]

\item The line $y = \sqrt{\frac{a}{b}} x$ meets $P_2$ at the point $D$.

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]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2019 Q8 [9]}}

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