A-LevelMathematicsDifferentiation9709/31 · May/June 2023 · Q58 Marks

The equation of a curve is x²y – ay² = 4a³, where a is a non-zero constant.

(a)Show that dy/dx = 2xy / (2ay – x²)[4]
(b)Hence find the coordinates of the points where the tangent to the curve is parallel to the y-axis.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

[B1] State or imply 2xy + x² dy/dx as derivative of x²y Accept partial: ∂/∂x (x²y) → 2xy.
[B1] State or imply -2ay dy/dx as derivative of -ay² Accept partial: ∂/∂y (-ay²) → -2ay.
[M1] Equate attempted derivative to zero and solve for dy/dx
[A1] Obtain answer dy/dx = 2xy / (2ay – x²) from full and correct working AG
[*M1] State or imply 2ay – x² = 0
[DM1] Substitute into equation of curve to obtain equation in x and a or in y and a e.g. x⁴/(2a) - x⁴/(4a) = 4a³ (from x^2(x^2/2a) - a(x^2/2a)^2 = 4a^3)
[A1] Obtain one correct point e.g. (2a, 2a).
[A1] Obtain second correct point and no others e.g. (-2a, 2a).

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Topic: Differentiation, A-Level Mathematics (syllabus 9709). Structured question, 8 marks.
Source: A-Level Mathematics, May/June 2023 Paper 3 Variant 1, question 5.

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