Standard +0.8 This is a Further Maths question requiring students to use the discriminant condition for tangency (b²-4ac=0) by substituting the line into the ellipse equation. While the algebraic manipulation is straightforward once the method is identified, recognizing that tangency means exactly one solution and applying the discriminant to a quadratic in x (or y) requires solid understanding beyond routine procedures. The 'without differentiation' constraint guides students toward the correct approach, but this is still above-average difficulty for requiring problem-solving insight rather than just applying a standard algorithm.
The ellipse \(E\) has equation
$$x^2 + \frac{y^2}{9} = 1$$
The line with equation \(y = mx + 4\) is a tangent to \(E\)
Without using differentiation show that \(m = \pm\sqrt{7}\)
[4 marks]
Question 8:
8 | Forms an equation by
eliminating y or x | 1.1a | M1 | (mx+4)2
x2 + =1
9
9x2 + m2x2 + 8mx + 16 = 9
(9 + m2)x2 + 8mx + 7 = 0
For a tangent, b 2 − 4 a c = 0
64m2 – 28(9 + m2) = 0
36m2 = 252
m2 = 7
m = ± 7
Obtains a correct three-term
quadratic equation.
PI correct discriminant | 1.1b | A1
Forms a quadratic equation in m
or m2 using b2 −4ac=0 | 2.2a | M1
Completes fully correct working
to obtain m = ± 7
AG | 2.1 | R1
Question total | 4
Q | Marking instructions | AO | Marks | Typical solution
The ellipse $E$ has equation
$$x^2 + \frac{y^2}{9} = 1$$
The line with equation $y = mx + 4$ is a tangent to $E$
Without using differentiation show that $m = \pm\sqrt{7}$
[4 marks]
\hfill \mbox{\textit{AQA Further Paper 1 2024 Q8 [4]}}