June 2025 Paper 1 Q8

OCR MEICurrent spec7 marksCo-ordinate Geometry Quadratics

(a) Determine the two values of \(k\) for which the line \(y = 3x + 5\) is a tangent to the curve \(y = k - kx - x^2\). [4]

(b) These values of \(k\) are used to define two curves of the form \(y = k - kx - x^2\).

Determine the coordinates of the point of intersection of these two curves. [3]

Mark scheme (a)

Scheme	Marks	AO
Curve intersects \(y = 3x+5\) when \(3x + 5 = k - kx - x^2\)	M1	3.1a
\(x^2 + x(k+3) + (5-k) = 0\)	M1	1.1
Tangent when equal roots so discriminant \(=0\) \((k+3)^2 - 4(5-k) = 0\)	M1	1.1a
\(k^2 + 10k - 11 = 0\) Giving \(k = -11,\ 1\)	A1	1.1
	[4]

Notes

M1: Attempts to solve simultaneously

M1: Collects terms. Must see \(=0\).
May be implied by correct discriminant

M1: Forms discriminant of their equation and equates to zero

A1: cao

Alternative method

Scheme	Marks
For curve \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -k - 2x = 3\) at point of contact	M1
\(3x + 5 = k - kx - x^2\)	M1
Either substitute \(x = -\left(\dfrac{k+3}{2}\right)\) leading to \(k^2 + 10k - 11 = 0\)	M1
Giving \(k = -11,\ 1\)	A1
Or substitute \(k = -(2x+3)\) leading to \(x^2 - 2x - 8 = 0\)	M1
Giving \(x = -2,\ 4\) which gives \(k = 1,\ -11\)	A1

M1: Uses gradient of 3 to form an equation linking \(k\) and \(x\)

M1: Forms second equation for the point of intersection oe

M1: (Either) Uses above quadratic equation and an expression for \(x\) obtained from gradient to form a quadratic in \(k\) only

A1: cao

M1: (Or) Uses above quadratic equation and an expression for \(k\) to form a quadratic in \(x\) only, leading to values for \(k\)

A1: cao

Mark scheme (b)

Scheme	Marks	AO
Intersect when \(-11 + 11x - x^2 = 1 - x - x^2\)	M1	2.1
\(x = 1\)	A1	1.1
\((1,\ -1)\)	A1	2.1
	[3]

Notes

M1: Uses their values for \(k\) and attempts to solve simultaneously

A1: Note – using any values of \(k\) will give the same point of intersection for full credit