Edexcel F3 2018 Specimen — Question 2 11 marks

Exam BoardEdexcel
ModuleF3 (Further Pure Mathematics 3)
Year2018
SessionSpecimen
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConic sections
TypeEllipse tangent/normal equation derivation
DifficultyChallenging +1.2 This is a Further Maths conic sections question requiring parametric representation, implicit differentiation for the normal, and coordinate geometry for area calculation. While it involves multiple steps (11 marks total), the techniques are standard for F3: finding the normal using dy/dx from parametric form, finding intersection points, and computing triangle area. The algebraic manipulation is moderately involved but follows predictable patterns for this topic.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation
An ellipse has equation $$\frac{x^2}{25} + \frac{y^2}{4} = 1$$ The point \(P\) lies on the ellipse and has coordinates \((5\cos \theta, 2\sin \theta)\), \(0 < \theta < \frac{\pi}{2}\) The line \(L\) is a normal to the ellipse at the point \(P\).
  1. Show that an equation for \(L\) is $$5x \sin \theta - 2y \cos \theta = 21 \sin \theta \cos \theta$$ [5]
Given that the line \(L\) crosses the \(y\)-axis at the point \(Q\) and that \(M\) is the midpoint of \(PQ\),
  1. find the exact area of triangle \(OPM\), where \(O\) is the origin, giving your answer as a multiple of \(\sin 2\theta\) [6]
Question 2:
AnswerMarks
2(a)x2 y2
(cid:14) (cid:32)1, P(5cos(cid:84),2sin(cid:84))
25 4
dx dy
(cid:32)(cid:16)5sin(cid:84), (cid:32)2cos(cid:84)
d(cid:84) d(cid:84)
or
2x 2y dy
(cid:14) (cid:32)0
AnswerMarks
25 4 dxCorrect derivatives or correct
implicit differentiationB1
dy 2cos(cid:84)
(cid:32)
AnswerMarks
dx (cid:16)5sin(cid:84)Divides their derivatives
correctly or substitutes and
AnswerMarks
rearrangesM1
5sin(cid:84)
M (cid:32)
AnswerMarks
N 2cos(cid:84)Correct perpendicular gradient
ruleM1
5sin(cid:84)
y(cid:16)2sin(cid:84)(cid:32) (cid:11)x(cid:16)5cos(cid:84)(cid:12)
AnswerMarks
2cos(cid:84)Correct straight line method (any
complete method) Must use
AnswerMarks Guidance
their gradient of the normal.M1
5xsin(cid:84)(cid:16)2ycos(cid:84)(cid:32)21sin(cid:84)cos(cid:84)*cso A1*
(5)
AnswerMarks
(b)21
At Q, x = 0 (cid:159) y(cid:32)(cid:16) sin(cid:84)
AnswerMarks
2B1
(cid:167)0(cid:14)5cos(cid:84) 2sin(cid:84)(cid:16) 21sin(cid:84)(cid:183)
M is , 2
(cid:168) (cid:184)
(cid:169) 2 2 (cid:185)
(cid:167) (cid:167)5 17 (cid:183)(cid:183)
(cid:32) cos(cid:84), (cid:16) sin(cid:84)
(cid:168) (cid:168) (cid:184)(cid:184)
AnswerMarks
(cid:169) (cid:169)2 4 (cid:185)(cid:185)Correct mid-point method for at
least one coordinate
Can be implied by a correct x
AnswerMarks
coordinateM1
21
Lcutsx-axisat cos(cid:84)
AnswerMarks
5B1
Area OPM = OLP
+OLM
1 21 1 21 17
. cos(cid:84).2sin(cid:84)(cid:14) . cos(cid:84). sin(cid:84)
AnswerMarks
2 5 2 5 4M1: Correct triangle area
method using their coordinatesM1 A1
A1: Correct expression
105
(cid:32) sin2(cid:84)
AnswerMarks
16Or 6.5625sin2(cid:84) must be
positiveA1
(6)
AnswerMarks Guidance
QuestionScheme Marks
2(b)
AnswerMarks
continuedAlternative 1: Using Area OPM
See above for B1M1B1 M1
1 0 5cos(cid:84) 5cos(cid:84) 0
Area (cid:39)OPM (cid:32) 2
2 0 2sin(cid:84) (cid:16)17sin(cid:84) 0
AnswerMarks
4M1: Correct determinant
with their coords, with 2 or
0
3 points. should be at
0
both or neither end.
A1: Correct determinant
(There are more
complicated determinants
AnswerMarks
using the 3 points.)M1 A1
1(cid:167) 85 (cid:183)
(cid:32) 0(cid:14)5sin(cid:84)cos(cid:84)(cid:14)0(cid:16)0(cid:14) sin(cid:84)cos(cid:84)(cid:16)0
(cid:168) (cid:184)
AnswerMarks Guidance
2(cid:169) 4 (cid:185)A1 A1
105
(cid:32) sin(cid:84)cos(cid:84)
4
105
(cid:32) sin2(cid:84)
AnswerMarks
16A1
(6)
Alternative 2: Using Area OPQ
21
At Q, x = 0 (cid:159) y(cid:32)(cid:16) sin(cid:84)
AnswerMarks
2B1
1 5cos(cid:84) 0
Area (cid:39)OPQ(cid:32)
2 2sin(cid:84) (cid:16)21sin(cid:84)
AnswerMarks
2Can be implied by the
following lineM1 A1
1 105
(cid:32) (cid:117) sin(cid:84)cos(cid:84)
AnswerMarks
2 2OQ is base, x coord of P is
heightA1
105
(cid:32) sin2(cid:84)
8
1
Area OPM = Area OPQ
AnswerMarks
2M1
105
(cid:32) sin2(cid:84)
AnswerMarks
16A1
(6)
AnswerMarks Guidance
QuestionScheme Marks
2(b)
AnswerMarks
continuedAlternative 3
21
At Q, x = 0 (cid:159) y(cid:32)(cid:16) sin(cid:84)
AnswerMarks
2B1
(cid:167)0(cid:14)5cos(cid:84) 2sin(cid:84)(cid:16)21sin(cid:84)(cid:183) (cid:167) (cid:167)5 17 (cid:183)(cid:183)
M is , 2 (cid:32) cos(cid:84), (cid:16) sin(cid:84)
(cid:168) (cid:184) (cid:168) (cid:168) (cid:184)(cid:184)
AnswerMarks
(cid:169) 2 2 (cid:185) (cid:169) (cid:169)2 4 (cid:185)(cid:185)M1
(cid:11) (cid:12)
AnswerMarks
OP(cid:32) 4sin2(cid:84)(cid:14)25cos2(cid:84) (cid:32) 4(cid:14)21cos2(cid:84)B1
5 2sin(cid:84) 17 21
cos(cid:84)(cid:117) (cid:14) sin(cid:84) sin(cid:84)
2 5cos(cid:84) 4 4
d (cid:32) (cid:32)
4sin2(cid:84) 4(cid:14)21cos2(cid:84)
(cid:14)1
25cos2(cid:84) 25cos2(cid:84)
21
sin(cid:84)
1
4
Area (cid:32) (cid:117) (cid:117) 4(cid:14)21cos2(cid:84)
2 4(cid:14)21cos2(cid:84)
AnswerMarks
25cos2(cid:84)M1 A1
105
(cid:32) sin2(cid:84)
AnswerMarks
16A1
(6)
(11 marks)
AnswerMarks Guidance
QuestionScheme Marks 
Question 2:
--- 2(a) ---
2(a) | x2 y2
(cid:14) (cid:32)1, P(5cos(cid:84),2sin(cid:84))
25 4
dx dy
(cid:32)(cid:16)5sin(cid:84), (cid:32)2cos(cid:84)
d(cid:84) d(cid:84)
or
2x 2y dy
(cid:14) (cid:32)0
25 4 dx | Correct derivatives or correct
implicit differentiation | B1
dy 2cos(cid:84)
(cid:32)
dx (cid:16)5sin(cid:84) | Divides their derivatives
correctly or substitutes and
rearranges | M1
5sin(cid:84)
M (cid:32)
N 2cos(cid:84) | Correct perpendicular gradient
rule | M1
5sin(cid:84)
y(cid:16)2sin(cid:84)(cid:32) (cid:11)x(cid:16)5cos(cid:84)(cid:12)
2cos(cid:84) | Correct straight line method (any
complete method) Must use
their gradient of the normal. | M1
5xsin(cid:84)(cid:16)2ycos(cid:84)(cid:32)21sin(cid:84)cos(cid:84)* | cso | A1*
(5)
(b) | 21
At Q, x = 0 (cid:159) y(cid:32)(cid:16) sin(cid:84)
2 | B1
(cid:167)0(cid:14)5cos(cid:84) 2sin(cid:84)(cid:16) 21sin(cid:84)(cid:183)
M is , 2
(cid:168) (cid:184)
(cid:169) 2 2 (cid:185)
(cid:167) (cid:167)5 17 (cid:183)(cid:183)
(cid:32) cos(cid:84), (cid:16) sin(cid:84)
(cid:168) (cid:168) (cid:184)(cid:184)
(cid:169) (cid:169)2 4 (cid:185)(cid:185) | Correct mid-point method for at
least one coordinate
Can be implied by a correct x
coordinate | M1
21
Lcutsx-axisat cos(cid:84)
5 | B1
Area OPM = OLP
+OLM
1 21 1 21 17
. cos(cid:84).2sin(cid:84)(cid:14) . cos(cid:84). sin(cid:84)
2 5 2 5 4 | M1: Correct triangle area
method using their coordinates | M1 A1
A1: Correct expression
105
(cid:32) sin2(cid:84)
16 | Or 6.5625sin2(cid:84) must be
positive | A1
(6)
Question | Scheme | Marks
2(b)
continued | Alternative 1: Using Area OPM
See above for B1M1 | B1 M1
1 0 5cos(cid:84) 5cos(cid:84) 0
Area (cid:39)OPM (cid:32) 2
2 0 2sin(cid:84) (cid:16)17sin(cid:84) 0
4 | M1: Correct determinant
with their coords, with 2 or
0
3 points. should be at
0
both or neither end.
A1: Correct determinant
(There are more
complicated determinants
using the 3 points.) | M1 A1
1(cid:167) 85 (cid:183)
(cid:32) 0(cid:14)5sin(cid:84)cos(cid:84)(cid:14)0(cid:16)0(cid:14) sin(cid:84)cos(cid:84)(cid:16)0
(cid:168) (cid:184)
2(cid:169) 4 (cid:185) | A1 | A1
105
(cid:32) sin(cid:84)cos(cid:84)
4
105
(cid:32) sin2(cid:84)
16 | A1
(6)
Alternative 2: Using Area OPQ
21
At Q, x = 0 (cid:159) y(cid:32)(cid:16) sin(cid:84)
2 | B1
1 5cos(cid:84) 0
Area (cid:39)OPQ(cid:32)
2 2sin(cid:84) (cid:16)21sin(cid:84)
2 | Can be implied by the
following line | M1 A1
1 105
(cid:32) (cid:117) sin(cid:84)cos(cid:84)
2 2 | OQ is base, x coord of P is
height | A1
105
(cid:32) sin2(cid:84)
8
1
Area OPM = Area OPQ
2 | M1
105
(cid:32) sin2(cid:84)
16 | A1
(6)
Question | Scheme | Marks
2(b)
continued | Alternative 3
21
At Q, x = 0 (cid:159) y(cid:32)(cid:16) sin(cid:84)
2 | B1
(cid:167)0(cid:14)5cos(cid:84) 2sin(cid:84)(cid:16)21sin(cid:84)(cid:183) (cid:167) (cid:167)5 17 (cid:183)(cid:183)
M is , 2 (cid:32) cos(cid:84), (cid:16) sin(cid:84)
(cid:168) (cid:184) (cid:168) (cid:168) (cid:184)(cid:184)
(cid:169) 2 2 (cid:185) (cid:169) (cid:169)2 4 (cid:185)(cid:185) | M1
(cid:11) (cid:12)
OP(cid:32) 4sin2(cid:84)(cid:14)25cos2(cid:84) (cid:32) 4(cid:14)21cos2(cid:84) | B1
5 2sin(cid:84) 17 21
cos(cid:84)(cid:117) (cid:14) sin(cid:84) sin(cid:84)
2 5cos(cid:84) 4 4
d (cid:32) (cid:32)
4sin2(cid:84) 4(cid:14)21cos2(cid:84)
(cid:14)1
25cos2(cid:84) 25cos2(cid:84)
21
sin(cid:84)
1
4
Area (cid:32) (cid:117) (cid:117) 4(cid:14)21cos2(cid:84)
2 4(cid:14)21cos2(cid:84)
25cos2(cid:84) | M1 A1
105
(cid:32) sin2(cid:84)
16 | A1
(6)
(11 marks)
Question | Scheme | Marks
An ellipse has equation
$$\frac{x^2}{25} + \frac{y^2}{4} = 1$$

The point $P$ lies on the ellipse and has coordinates $(5\cos \theta, 2\sin \theta)$, $0 < \theta < \frac{\pi}{2}$

The line $L$ is a normal to the ellipse at the point $P$.

\begin{enumerate}[label=(\alph*)]
\item Show that an equation for $L$ is
$$5x \sin \theta - 2y \cos \theta = 21 \sin \theta \cos \theta$$
[5]
\end{enumerate}

Given that the line $L$ crosses the $y$-axis at the point $Q$ and that $M$ is the midpoint of $PQ$,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the exact area of triangle $OPM$, where $O$ is the origin, giving your answer as a multiple of $\sin 2\theta$
[6]
\end{enumerate}

\hfill \mbox{\textit{Edexcel F3 2018 Q2 [11]}}
This paper (8 questions)
View full paper
Q1 6 Q2 11 Q3 12 Q4 9 Q5 7 Q6 9 Q7 11 Q8 10