C1 June 2013 (R) Q11

EdexcelOld spec9 marksCo-ordinate Geometry Quadratics

11.

Figure 2: the curve x^2 + 4y^2 - 2x = 35, an oval about the x-axis, and the line y = x + 2 meeting it at A in the first quadrant and B in the third quadrant — Figure 2

The line \(y = x + 2\) meets the curve \(x^2 + 4y^2 - 2x = 35\) at the points \(A\) and \(B\) as shown in Figure 2.

(a) Find the coordinates of \(A\) and the coordinates of \(B\). (6)

(b) Find the distance \(AB\) in the form \(r\sqrt{2}\) where \(r\) is a rational number. (3)

Mark scheme (a)

Scheme	Marks
\(y = x + 2 \Rightarrow x^2 + 4(x + 2)^2 - 2x = 35\)	M1
Alternative: \(\dfrac{2x - x^2 + 35}{4} = (x + 2)^2\) or \(\sqrt{\dfrac{2x - x^2 + 35}{4}} = (x + 2)\)
\(5x^2 + 14x - 19 = 0\)	M1
\((5x + 19)(x - 1) = 0 \Rightarrow x = ..\)	dM1
\(x = -\dfrac{19}{5},\ x = 1\)	A1 for both
\(y = -\dfrac{9}{5},\ y = 3\)	M1
Coordinates are \(\left(-\dfrac{19}{5}, -\dfrac{9}{5}\right)\) and \((1, 3)\)	A1
	(6)

Notes

M1: Substitute \(y = \pm x \pm 2\) into \(x^2 + 4y^2 - 2x = 35\) to obtain an equation in \(x\) only.

M1: Multiply out and collects terms producing 3 term quadratic in any form.

dM1: Solves their quadratic, usual rules, as far as \(x\) = ... Dependent on the first M1 i.e. a correct method for eliminating \(y\) (or \(x\) – see below)

A1: Both correct

M1: Substitutes back into either given equation to find a value for \(y\)

A1: Correct matching pairs. Coordinates need not be given explicitly but it must be clear which \(x\) goes with which \(y\)

Alternative to part (a)

Scheme	Marks
\(x = y - 2 \Rightarrow (y - 2)^2 + 4y^2 - 2(y - 2) =\)	M1
\(5y^2 - 6y - 27 = 0\)	M1
\((5y + 9)(y - 3) = 0 \Rightarrow y = ..\)	dM1
\(y = -\dfrac{9}{5},\ y = 3\)	A1 for both
\(x = -\dfrac{19}{5},\ x = 1\)	M1
Coordinates are \(\left(-\dfrac{19}{5}, -\dfrac{9}{5}\right)\) and \((1,3)\)	A1

M1: Substitutes \(x = \pm y \pm 2\) into \(x^2 + 4y^2 - 2x = 35\)

M1: Multiply out, collect terms producing 3 term quadratic in any form.

dM1: Solves their quadratic, usual rules, as far as \(y\) = ... Dependent on the first M1 i.e. a correct method for eliminating \(x\)

A1: Both correct

M1: Substitutes back into either given equation to find a value for \(x\)

A1: Correct matching pairs as above.

Mark scheme (b)

Scheme	Marks
\(d^2 = \left(1 - -\dfrac{19}{5}\right)^2 + \left(3 - -\dfrac{9}{5}\right)^2\) or \(d = \sqrt{\left(1 - -\dfrac{19}{5}\right)^2 + \left(3 - -\dfrac{9}{5}\right)^2}\)	M1A1ft
\(d = \dfrac{24}{5}\sqrt{2}\)	A1cao
	(3)
	(9 marks)

Notes

M1: Use of \(d^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2\) or \(d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\) where neither \((x_1 - x_2)\) nor \((y_1 - y_2)\) are zero.

A1ft: Correct ft expression for \(d\) or \(d^2\) (may be un-simplified)

A1cao: Allow \(4.8\sqrt{2}\)