Question 6 - A-Level Maths

AQA Further Paper 2 2021 June — Question 6 8 marks

Exam Board	AQA
Module	Further Paper 2 (Further Paper 2)
Year	2021
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Conic sections
Type	Conic translation and transformation
Difficulty	Challenging +1.2 This is a multi-part Further Maths question on conic sections that requires understanding of translations, transformations, and tangent properties. Parts (a) and (b) are routine applications of transformations. Part (c) requires visualization and sketching skills. Part (d) requires geometric insight about symmetry and common tangents, but the reasoning is accessible once the setup is understood. While it's Further Maths content (inherently harder), the individual steps are relatively straightforward with no complex algebra or proof required.
Spec	1.02n Sketch curves: simple equations including polynomials 1.02w Graph transformations: simple transformations of f(x)1.02x Combinations of transformations: multiple transformations 1.03d Circles: equation (x-a)^2+(y-b)^2=r^2

6 The ellipse $E _ { 1 }$ has equation $$x ^ { 2 } + \frac { y ^ { 2 } } { 4 } = 1$$ $E _ { 1 }$ is translated by the vector $\left[ \begin{array} { l } 3 \\ 0 \end{array} \right]$ to give the ellipse $E _ { 2 }$ 6

Write down the equation of $E _ { 2 }$ 6
The ellipse $E _ { 3 }$ has equation $$\frac { x ^ { 2 } } { 4 } + ( y - 3 ) ^ { 2 } = 1$$ Describe the transformation that maps $E _ { 2 }$ to $E _ { 3 }$ 6
Each of the lines $L _ { A }$ and $L _ { B }$ is a tangent to both $E _ { 2 }$ and $E _ { 3 }$ $L _ { A }$ is closer to the origin than $L _ { B }$ $E _ { 2 }$ and $E _ { 3 }$ both lie between $L _ { A }$ and $L _ { B }$ Sketch and label $E _ { 2 } , E _ { 3 } , L _ { A }$ and $L _ { B }$ on the axes below.
You do not need to show the values of the axis intercepts for $L _ { A }$ and $L _ { B }$ \includegraphics[max width=\textwidth, alt={}, center]{13abb93f-2fef-465c-980c-3b412de06618-09_1095_1095_726_475} 6
Explain, without doing any calculations, why $L _ { A }$ has an equation of the form $$x + y = c$$ where $c$ is a constant.

Show mark scheme Show mark scheme source

Question 6(a):

Answer	Marks	Guidance
$(x-3)^2 + \frac{y^2}{4} = 1$	B1	Correct equation of $E_2$

Question 6(b):

Answer	Marks	Guidance
Reflection in the line $y = x$	B1	Condone a correct sequence of transformations

Question 6(c):

Answer	Marks
Two ellipses, one crossing positive $x$-axis and other crossing positive $y$-axis	B1
Correct axis intercepts shown for both ellipses	B1
At least one correct tangent drawn; condone $x=2$ or $y=2$	B1
Both lines drawn and labelled correctly	B1

Question 6(d):

Answer	Marks	Guidance
Points on $E_2$ and $E_3$ joined by $L_A$ are symmetrical about $y=x$, therefore line is perpendicular to $y=x$	E1	Uses fact that $E_3$ is reflection of $E_2$ in line $y=x$
Tangent is perpendicular to $y=x$, has gradient $-1$, and is of the form $y = -x + c$, so equation is $x + y = c$	E1	Explains tangent is perpendicular to $y=x$ and concludes equation is $x+y=c$

## Question 6(a):

$(x-3)^2 + \frac{y^2}{4} = 1$ | B1 | Correct equation of $E_2$

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## Question 6(b):

Reflection in the line $y = x$ | B1 | Condone a correct sequence of transformations

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## Question 6(c):

Two ellipses, one crossing positive $x$-axis and other crossing positive $y$-axis | B1 |

Correct axis intercepts shown for both ellipses | B1 |

At least one correct tangent drawn; condone $x=2$ or $y=2$ | B1 |

Both lines drawn and labelled correctly | B1 |

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## Question 6(d):

Points on $E_2$ and $E_3$ joined by $L_A$ are symmetrical about $y=x$, therefore line is perpendicular to $y=x$ | E1 | Uses fact that $E_3$ is reflection of $E_2$ in line $y=x$

Tangent is perpendicular to $y=x$, has gradient $-1$, and is of the form $y = -x + c$, so equation is $x + y = c$ | E1 | Explains tangent is perpendicular to $y=x$ and concludes equation is $x+y=c$

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Show LaTeX source

6 The ellipse $E _ { 1 }$ has equation

$$x ^ { 2 } + \frac { y ^ { 2 } } { 4 } = 1$$

$E _ { 1 }$ is translated by the vector $\left[ \begin{array} { l } 3 \\ 0 \end{array} \right]$ to give the ellipse $E _ { 2 }$\\
6
\begin{enumerate}[label=(\alph*)]
\item Write down the equation of $E _ { 2 }$

6
\item The ellipse $E _ { 3 }$ has equation

$$\frac { x ^ { 2 } } { 4 } + ( y - 3 ) ^ { 2 } = 1$$

Describe the transformation that maps $E _ { 2 }$ to $E _ { 3 }$

6
\item Each of the lines $L _ { A }$ and $L _ { B }$ is a tangent to both $E _ { 2 }$ and $E _ { 3 }$\\
$L _ { A }$ is closer to the origin than $L _ { B }$\\
$E _ { 2 }$ and $E _ { 3 }$ both lie between $L _ { A }$ and $L _ { B }$\\
Sketch and label $E _ { 2 } , E _ { 3 } , L _ { A }$ and $L _ { B }$ on the axes below.\\
You do not need to show the values of the axis intercepts for $L _ { A }$ and $L _ { B }$\\
\includegraphics[max width=\textwidth, alt={}, center]{13abb93f-2fef-465c-980c-3b412de06618-09_1095_1095_726_475}

6
\item Explain, without doing any calculations, why $L _ { A }$ has an equation of the form

$$x + y = c$$

where $c$ is a constant.
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2021 Q6 [8]}}

This paper (13 questions)

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Q1 1 Q2 1 Q3 1 Q4 7 Q5 5 Q6 8 Q7 7 Q8 6 Q9 14 Q10 13 Q11 9 Q12 12 Q13 16

Answer	Marks
Two ellipses, one crossing positive \(x\)-axis and other crossing positive \(y\)-axis	B1
Correct axis intercepts shown for both ellipses	B1
At least one correct tangent drawn; condone \(x=2\) or \(y=2\)	B1
Both lines drawn and labelled correctly	B1

Answer	Marks	Guidance
Points on \(E_2\) and \(E_3\) joined by \(L_A\) are symmetrical about \(y=x\), therefore line is perpendicular to \(y=x\)	E1	Uses fact that \(E_3\) is reflection of \(E_2\) in line \(y=x\)
Tangent is perpendicular to \(y=x\), has gradient \(-1\), and is of the form \(y = -x + c\), so equation is \(x + y = c\)	E1	Explains tangent is perpendicular to \(y=x\) and concludes equation is \(x+y=c\)