Question 5 - A-Level Maths

OCR MEI FP2 2013 January — Question 5 18 marks

Exam Board	OCR MEI
Module	FP2 (Further Pure Mathematics 2)
Year	2013
Session	January
Marks	18
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Polar coordinates
Type	Sketch polar curve
Difficulty	Challenging +1.8 This is a substantial Further Maths polar curves question requiring multiple sketches, conversion to Cartesian form, and algebraic analysis of domains. While the individual techniques (sketching polar curves, coordinate conversion, domain analysis) are standard FP2 material, the multi-part structure, need to identify geometric features across multiple parameter values, and the algebraic manipulation in parts (v)-(vi) require sustained reasoning and good conceptual understanding of polar coordinates beyond routine exercises.
Spec	4.09b Sketch polar curves: r = f(theta)

5 This question concerns the curves with polar equation $$r = \sec \theta + a \cos \theta ,$$ where $a$ is a constant which may take any real value, and $0 \leqslant \theta \leqslant 2 \pi$.

On a single diagram, sketch the curves for $a = 0 , a = 1 , a = 2$.
On a single diagram, sketch the curves for $a = 0 , a = - 1 , a = - 2$.
Identify a feature that the curves for $a = 1 , a = 2 , a = - 1 , a = - 2$ share.
Name a distinctive feature of the curve for $a = - 1$, and a different distinctive feature of the curve for $a = - 2$.
Show that, in cartesian coordinates, equation (*) may be written $$y ^ { 2 } = \frac { a x ^ { 2 } } { x - 1 } - x ^ { 2 }$$ Hence comment further on the feature you identified in part (iii).
Show algebraically that, when $a > 0$, the curve exists for $1 < x < 1 + a$. Find the set of values of $x$ for which the curve exists when $a < 0$.

Show mark scheme Show mark scheme source

Question 5:

Part (i):

Answer	Marks	Guidance
Answer	Marks	Guidance
Three curves of correct shape, correctly identified	B2, B1	Give B1 for two correct curves; $a=0$, $a=1$, $a=2$ from left to right
[3]

Part (ii):

Answer	Marks	Guidance
Answer	Marks	Guidance
Curve for $a = -1$	B1	Curve with cusp
Curve for $a = -2$	B1	Curve with loop
[2]

Part (iii):

Answer	Marks	Guidance
Answer	Marks	Guidance
Asymptote	B1
[1]

Part (iv):

Answer	Marks	Guidance
Answer	Marks	Guidance
$a = -1$: cusp	B1
$a = -2$: loop	B1
[2]

Part (v):

Answer	Marks	Guidance
Answer	Marks	Guidance
$r = \sec\theta + a\cos\theta \Rightarrow r\cos\theta = 1 + a\cos^2\theta$
$\Rightarrow x = 1 + a\left(\frac{x^2}{r^2}\right)$	M1	Using $x = r\cos\theta$
$\Rightarrow x - 1 = a\left(\frac{x^2}{x^2+y^2}\right)$	M1	Using $r^2 = x^2 + y^2$
$\Rightarrow x^2 + y^2 = a\left(\frac{x^2}{x-1}\right) \Rightarrow y^2 = a\left(\frac{x^2}{x-1}\right) - x^2$	M1, A1(ag)	Making $y^2$ subject
Hence asymptote at $x = 1$	B1
[5]

Part (vi):

Answer	Marks	Guidance
Answer	Marks	Guidance
Curve exists for $y^2 \geq 0 \Rightarrow a\left(\frac{1}{x-1}\right) - 1 \geq 0$	M1	Considering $y^2 \geq 0$
If $a > 0$ then $x - 1 > 0$ and so $a \geq x-1$, i.e. $1 < x \leq 1+a$	M1, A1(ag)
If $a < 0$ then $x - 1 < 0$ and so $a \leq x-1$, i.e. $1+a \leq x < 1$	M1, A1
[5]

# Question 5:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Three curves of correct shape, correctly identified | B2, B1 | Give B1 for two correct curves; $a=0$, $a=1$, $a=2$ from left to right |
| **[3]** | | |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Curve for $a = -1$ | B1 | Curve with cusp |
| Curve for $a = -2$ | B1 | Curve with loop |
| **[2]** | | |

## Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Asymptote | B1 | |
| **[1]** | | |

## Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = -1$: cusp | B1 | |
| $a = -2$: loop | B1 | |
| **[2]** | | |

## Part (v):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = \sec\theta + a\cos\theta \Rightarrow r\cos\theta = 1 + a\cos^2\theta$ | | |
| $\Rightarrow x = 1 + a\left(\frac{x^2}{r^2}\right)$ | M1 | Using $x = r\cos\theta$ |
| $\Rightarrow x - 1 = a\left(\frac{x^2}{x^2+y^2}\right)$ | M1 | Using $r^2 = x^2 + y^2$ |
| $\Rightarrow x^2 + y^2 = a\left(\frac{x^2}{x-1}\right) \Rightarrow y^2 = a\left(\frac{x^2}{x-1}\right) - x^2$ | M1, A1(ag) | Making $y^2$ subject |
| Hence asymptote at $x = 1$ | B1 | |
| **[5]** | | |

## Part (vi):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Curve exists for $y^2 \geq 0 \Rightarrow a\left(\frac{1}{x-1}\right) - 1 \geq 0$ | M1 | Considering $y^2 \geq 0$ |
| If $a > 0$ then $x - 1 > 0$ and so $a \geq x-1$, i.e. $1 < x \leq 1+a$ | M1, A1(ag) | |
| If $a < 0$ then $x - 1 < 0$ and so $a \leq x-1$, i.e. $1+a \leq x < 1$ | M1, A1 | |
| **[5]** | | |

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5 This question concerns the curves with polar equation

$$r = \sec \theta + a \cos \theta ,$$

where $a$ is a constant which may take any real value, and $0 \leqslant \theta \leqslant 2 \pi$.\\
(i) On a single diagram, sketch the curves for $a = 0 , a = 1 , a = 2$.\\
(ii) On a single diagram, sketch the curves for $a = 0 , a = - 1 , a = - 2$.\\
(iii) Identify a feature that the curves for $a = 1 , a = 2 , a = - 1 , a = - 2$ share.\\
(iv) Name a distinctive feature of the curve for $a = - 1$, and a different distinctive feature of the curve for $a = - 2$.\\
(v) Show that, in cartesian coordinates, equation (*) may be written

$$y ^ { 2 } = \frac { a x ^ { 2 } } { x - 1 } - x ^ { 2 }$$

Hence comment further on the feature you identified in part (iii).\\
(vi) Show algebraically that, when $a > 0$, the curve exists for $1 < x < 1 + a$.

Find the set of values of $x$ for which the curve exists when $a < 0$.

\hfill \mbox{\textit{OCR MEI FP2 2013 Q5 [18]}}

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Q1 18 Q2 18 Q3 18 Q4 18 Q5 18