AS June 2025 Paper 1 Q15

AQACurrent spec11 marksPolar Coordinates

15 The graph \(G_1\) has the polar equation

\[r = \frac{8}{\cos\theta + \sqrt{3}\sin\theta} \qquad -\frac{\pi}{6} \lt \theta \lt \frac{5\pi}{6}\]

The graph \(G_2\) has the polar equation

\[\theta = \frac{2\pi}{3}\]

(a) Show that \(G_1\) is a straight line by writing it as a Cartesian equation in the form \(y = mx + c\) where \(m\) and \(c\) are constants to be found. [3 marks]

(b) On the initial line shown in Figure 1, sketch \(G_1\) and \(G_2\)

Include the polar coordinates of the intersection point of \(G_1\) with the initial line. [3 marks]

Figure 1: the pole O with the initial line drawn horizontally to the right — Figure 1

(d) Find the area of the region bounded by the initial line and graphs \(G_1\) and \(G_2\) [2 marks]

Mark scheme (a)

Scheme	Marks	AO
Rearranges the equation in terms of \(r\sin\theta\) and \(r\cos\theta\)	M1	3.1a
Replaces \(r\cos\theta\) with \(x\) and \(r\sin\theta\) with \(y\)	M1	1.2
Obtains \(y = -\dfrac{1}{\sqrt{3}}x + \dfrac{8}{\sqrt{3}}\)	R1	2.1
	(3)

Typical solution

\[r = \frac{8}{\cos\theta + \sqrt{3}\sin\theta}\]\[r\cos\theta + \sqrt{3}r\sin\theta = 8\]\[x + \sqrt{3}y = 8\]\[y = -\frac{1}{\sqrt{3}}x + \frac{8}{\sqrt{3}}\]

Sketch: G2 is a half-line from O into the second quadrant; G1 is a straight line crossing G2, passing through the second, first and fourth quadrants and meeting the initial line at (8, 0); O is labelled (0, 0) — Mark scheme (b)

Scheme	Marks	AO
Draws a straight line through the 2nd, 1st and 4th quadrants.	B1	1.1b
Gives the initial line intersection as (8, 0) or Writes 8 next to the initial line intersection.	B1	1.1b
Draws a straight line from \(O\) into the 2nd quadrant. Must intersect \(G_1\) if drawn.	B1	1.1b
	(3)

Mark scheme (c)

Scheme	Marks	AO
Substitutes \(\theta = \dfrac{2\pi}{3}\) into the RHS of the \(G_1\) polar equation.	M1	3.1a
Obtains 8	A1	1.1b
Obtains \(\left(8, \dfrac{2\pi}{3}\right)\)	A1	1.1b
	(3)

Typical solution

\[\begin{aligned} r &= \frac{8}{\cos\frac{2\pi}{3} + \sqrt{3}\sin\frac{2\pi}{3}} \\ &= 8\end{aligned}\]\[\text{Intersection point} = \left(8, \frac{2\pi}{3}\right)\]

Mark scheme (d)

Scheme	Marks	AO
Shows a full method for the area of the triangle. Accept \(\dfrac{1}{2} \times 8 \times 8 \times \sin 120\)	M1	3.1a
Obtains \(16\sqrt{3}\) Accept AWRT 27.7 FT their 8’s to at least 3sf if approximated	A1F	1.1b
	(2)
	(11 marks)

Typical solution

\[\begin{aligned}\text{area} &= \frac{1}{2} \times 8 \times 8 \times \sin\left(\frac{2\pi}{3}\right) \\ &= 16\sqrt{3}\end{aligned}\]