June 2022 Paper 2 Q10

OCR MEICurrent spec7 marksCo-ordinate Geometry Parametric Equations

10 The parametric equations of a curve are

\(x = 2 + 5\cos\theta\) and \(y = 1 + 5\sin\theta\), where \(0 \leqslant \theta \leqslant 2\pi\).

(a) Determine the cartesian equation of the curve. [3]

(b) Hence or otherwise, find the equation of the tangent to the curve at the point \((5, -3)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers to be determined. [4]

Mark scheme (a)

Scheme	Marks	AO
\((x - 2) = 5\cos\theta\) and \((y - 1) = 5\sin\theta\)	M1	3.1a
\((x - 2)^2 + (y - 1)^2 = (5\cos\theta)^2 + (5\sin\theta)^2\) oe	M1	1.1
\((x - 2)^2 + (y - 1)^2 = 5^2\) oe isw or \(\dfrac{(x-2)^2}{5^2} + \dfrac{(y-1)^2}{5^2} = 1\) oe isw	A1	1.1
	[3]

Notes

M1: allow sign errors

M1: or \(\left(\frac{x-2}{5}\right)^2 + \left(\frac{y-1}{5}\right)^2 = \cos^2\theta + \sin^2\theta\) oe

A1: may see eg \(x^2 - 4x + y^2 - 2y = 20\)
if M0M0 allow SC1 for \(y = 1 + 5\sin\left(\cos^{-1}\left(\frac{x-2}{5}\right)\right)\) or \(x = 2 + 5\cos\left(\sin^{-1}\left(\frac{y-1}{5}\right)\right)\)

Alternatively

Scheme	Marks
\(x^2 = (2 + 5\cos\theta)^2\) and \(y^2 = (1 + 5\sin\theta)^2\)	M1
\(x^2 + y^2 = 5 + 20\cos\theta + 10\sin\theta + 25\sin^2\theta + 25\cos^2\theta\)	M1
\(x^2 + y^2 = 20 + 4x + 2y\) oe isw	A1

M1: if only seen in expanded form, allow one coefficient error; allow sign errors

M1: must have terms in \(\cos\theta\) and \(\sin\theta\)

Alternatively

Scheme	Marks
radius = 5 and centre is (2, 1)	M1
\((x - 2)^2 + (y - 1)^2 = 5^2\)	M1 A1

M1: allow sign error in coordinates of centre

M1: FT their centre

A1: all correct

Mark scheme (b)

Scheme	Marks	AO
gradient of radius is \(\frac{-4}{3}\)	B1	3.1a
gradient of tangent is \(\frac{3}{4}\)	M1	2.1
\((y - -3) = \frac{3}{4}(x - 5)\) oe	M1	2.4
\(3x - 4y - 27 = 0\) or \(-3x + 4y + 27 = 0\)	A1	1.1
	[4]

Notes

M1: FT \(1 \div\) their \(-\frac{4}{3}\)

M1: allow one sign error; FT their \(\frac{3}{4}\)
may see \(-3 = \frac{3}{4} \times 5 + c\)

Alternatively

Scheme	Marks
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{5\cos\theta}{-5\sin\theta}\) oe	B1
substitution of \(\cos\theta = \frac{3}{5}\) and \(\sin\theta = -\frac{4}{5}\) oe or \((5, -3)\) in their \(\frac{\mathrm{d}y}{\mathrm{d}x}\)	M1
\((y - -3) = \frac{3}{4}(x - 5)\) oe	M1
\(3x - 4y - 27 = 0\) or \(-3x + 4y + 27 = 0\)	A1

B1: or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2-x}{y-1}\) oe
eg \(2(x - 2) + 2(y - 1)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)

M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3/5}{-(-4/5)}\) or \(\dfrac{2-5}{-3-1}\) oe; allow one sign error;

M1: allow one sign error; FT their \(\frac{3}{4}\)
may see \(-3 = \frac{3}{4} \times 5 + c\)