AQA Paper 2 2020 June — Question 8 10 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2020
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric curves and Cartesian conversion
TypeProperties of specific curves
DifficultyStandard +0.3 This is a straightforward parametric curves question requiring standard techniques: eliminating the parameter (trivial here since yΒ²=4tΒ²=4x), finding dy/dx using the chain rule, computing a gradient between two points, and applying the double angle formula. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation
The curve defined by the parametric equations $$x = t^2 \text{ and } y = 2t \quad -\sqrt{2} \leq t \leq \sqrt{2}$$ is shown in Figure 1 below. \includegraphics{figure_1}
  1. Find a Cartesian equation of the curve in the form \(y^2 = f(x)\) [2 marks]
  2. The point \(A\) lies on the curve where \(t = a\) The tangent to the curve at \(A\) is at an angle \(\theta\) to a line through \(A\) parallel to the \(x\)-axis. The point \(B\) has coordinates \((1, 0)\) The line \(AB\) is at an angle \(\phi\) to the \(x\)-axis. \includegraphics{figure_1_extended}
    1. By considering the gradient of the curve, show that $$\tan \theta = \frac{1}{a}$$ [3 marks]
    2. Find \(\tan \phi\) in terms of \(a\). [2 marks]
    3. Show that \(\tan 2\theta = \tan \phi\) [3 marks]
Question 8:
AnswerMarks Guidance
8(a)Eliminates t 1.1a
𝑦𝑦 𝑦𝑦²
2Β²= =𝑑𝑑 4 π‘₯π‘₯ = 4
Writes the Cartesian equation
AnswerMarks Guidance
in the required form1.1b A1
Subtotal2
AnswerMarks
8(b)(i)Differentiates both ,
𝑑𝑑π‘₯π‘₯
with at least one correct
𝑑𝑑𝑑𝑑 = 2𝑑𝑑
𝑑𝑑𝑦𝑦
O𝑑𝑑𝑑𝑑r = 2
Differentiates their Β² = 4 to
dy
obtain y = A 𝑦𝑦 π‘₯π‘₯
dx
Or rearranges and differentiates
y =2 x and obtains
dy βˆ’ 1
= Ax 2
dx
AnswerMarks Guidance
O E3.1a M1
,
𝑑𝑑π‘₯π‘₯ 𝑑𝑑𝑦𝑦
𝑑𝑑𝑑𝑑 = 2𝑑𝑑 𝑑𝑑𝑑𝑑 = 2
𝑑𝑑𝑦𝑦 2 1
𝑑𝑑π‘₯π‘₯ = 2π‘Žπ‘Ž = π‘Žπ‘Ž
The gradient of a line is equal to
the tangent of the angle between
the line and the horizontal hence
1
tanΞΈ=
a
AnswerMarks Guidance
Obtains correct at t = a1.1b A1
𝑑𝑑𝑦𝑦
Explains that the gradient of a
𝑑𝑑π‘₯π‘₯
line is the tangent of the angle
between the line and the
horizontal
or
shows on right-angled triangle
tanΞΈ
on diagram and links to
and
1
concludes tanΞΈ=
AnswerMarks Guidance
a2.4 E1
Subtotal3
AnswerMarks
8(b)(ii)Uses formula for gradient of
straight line with points A and B
Must have a2βˆ’1 or 1βˆ’a2 in
AnswerMarks Guidance
denominator1.1a M1
tanφ=
a2 βˆ’1
2a
=
a2 βˆ’1
Obtains correct tanφ
AnswerMarks Guidance
OE1.1b A1
Subtotal2
AnswerMarks Guidance
8(b)(iii)States double angle formula for
tan2ΞΈ1.2 B1
tan2ΞΈ=
1βˆ’tan2ΞΈ
1
2Γ—
a
=
2
1
1βˆ’
 
ο£­aο£Έ
2a
=
a2 βˆ’1
= tanφ
1
Substitutes tanΞΈ= into their
a
2tanΞΈ
tan2ΞΈ=
AnswerMarks Guidance
1Β±tan2ΞΈ1.1a M1
Simplifies and completes
argument to show required
AnswerMarks Guidance
result2.1 R1
Subtotal3
Question Total10
QMarking instructions AO 
Question 8:
--- 8(a) ---
8(a) | Eliminates t | 1.1a | M1 | ,
𝑦𝑦 𝑦𝑦²
2Β²= =𝑑𝑑 4 π‘₯π‘₯ = 4
Writes the Cartesian equation
in the required form | 1.1b | A1
Subtotal | 2
--- 8(b)(i) ---
8(b)(i) | Differentiates both ,
𝑑𝑑π‘₯π‘₯
with at least one correct
𝑑𝑑𝑑𝑑 = 2𝑑𝑑
𝑑𝑑𝑦𝑦
O𝑑𝑑𝑑𝑑r = 2
Differentiates their Β² = 4 to
dy
obtain y = A 𝑦𝑦 π‘₯π‘₯
dx
Or rearranges and differentiates
y =2 x and obtains
dy βˆ’ 1
= Ax 2
dx
O E | 3.1a | M1 | 𝑦𝑦 π‘₯π‘₯
,
𝑑𝑑π‘₯π‘₯ 𝑑𝑑𝑦𝑦
𝑑𝑑𝑑𝑑 = 2𝑑𝑑 𝑑𝑑𝑑𝑑 = 2
𝑑𝑑𝑦𝑦 2 1
𝑑𝑑π‘₯π‘₯ = 2π‘Žπ‘Ž = π‘Žπ‘Ž
The gradient of a line is equal to
the tangent of the angle between
the line and the horizontal hence
1
tanΞΈ=
a
Obtains correct at t = a | 1.1b | A1
𝑑𝑑𝑦𝑦
Explains that the gradient of a
𝑑𝑑π‘₯π‘₯
line is the tangent of the angle
between the line and the
horizontal
or
shows on right-angled triangle
tanΞΈ
on diagram and links to
and
1
concludes tanΞΈ=
a | 2.4 | E1
Subtotal | 3
--- 8(b)(ii) ---
8(b)(ii) | Uses formula for gradient of
straight line with points A and B
Must have a2βˆ’1 or 1βˆ’a2 in
denominator | 1.1a | M1 | 2aβˆ’0
tanφ=
a2 βˆ’1
2a
=
a2 βˆ’1
Obtains correct tanφ
OE | 1.1b | A1
Subtotal | 2
--- 8(b)(iii) ---
8(b)(iii) | States double angle formula for
tan2ΞΈ | 1.2 | B1 | 2tanΞΈ
tan2ΞΈ=
1βˆ’tan2ΞΈ
1
2Γ—
a
=
2
1
1βˆ’
 
ο£­aο£Έ
2a
=
a2 βˆ’1
= tanφ
1
Substitutes tanΞΈ= into their
a
2tanΞΈ
tan2ΞΈ=
1Β±tan2ΞΈ | 1.1a | M1
Simplifies and completes
argument to show required
result | 2.1 | R1
Subtotal | 3
Question Total | 10
Q | Marking instructions | AO | Marks | Typical solution
The curve defined by the parametric equations
$$x = t^2 \text{ and } y = 2t \quad -\sqrt{2} \leq t \leq \sqrt{2}$$

is shown in Figure 1 below.

\includegraphics{figure_1}

\begin{enumerate}[label=(\alph*)]
\item Find a Cartesian equation of the curve in the form $y^2 = f(x)$
[2 marks]

\item The point $A$ lies on the curve where $t = a$

The tangent to the curve at $A$ is at an angle $\theta$ to a line through $A$ parallel to the $x$-axis.

The point $B$ has coordinates $(1, 0)$

The line $AB$ is at an angle $\phi$ to the $x$-axis.

\includegraphics{figure_1_extended}

\begin{enumerate}[label=(\roman*)]
\item By considering the gradient of the curve, show that
$$\tan \theta = \frac{1}{a}$$
[3 marks]

\item Find $\tan \phi$ in terms of $a$.
[2 marks]

\item Show that $\tan 2\theta = \tan \phi$
[3 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2 2020 Q8 [10]}}
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