SUMMER 2026
PREDICTED PAPER
AQA A-level
MATHEMATICS
7357/1
Paper 1
Mark scheme
,1 (a) A curve C has parametric equations
x = 2t 3 y = 4t − 1 t >0
dy
Find an expression for in terms of t .
dx
Answer
Examiner Tips and Tricks
⎛⎜ dy ⎞⎟
⎜ ⎟
dy ⎝ dt ⎠
For parametric equations, =
dx ⎛⎜ dx ⎞⎟
⎜ ⎟
⎝ dt ⎠
Differentiate with respect to
Differentiate with respect to
[M1]
Divide
Examiner Tips and Tricks
dy
Make sure you divide in the correct order, with on the top!
dt
for more: tyrionpapers.com
, That fraction can be simplified
[M1 A1]
Examiner Tips and Tricks
dy 4
= 2 would also be a correct answer here, or you could also use laws of indices to
dx 6t
dy 2 −2
rewrite the answer as, for example, = t .
dx 3
(3 marks)
(b) Find the equation of the normal to C at the point ( 16, 7) .
Give your answer in the form ax + by + c = 0 where a , b and c are integers to be found.
Answer
Examiner Tips and Tricks
dy
The equation for is in terms of t , so in order to find the gradient of the tangent line
dx
you need to find the t value at the point ( 16, 7) .
At the point , , so
[M1]
Examiner Tips and Tricks
for more: tyrionpapers.com
, You can check your t value by substituting it into the x equation, to make sure it gives 16:
x = 2(2) 3 = 2(8) = 16 ✓
Find the gradient of the tangent by substituting into the formula for
[M1]
The gradient of the normal is the negative reciprocal of that
[A1]
Use the equation of a line , with and
Rearrange into the required form
[A1]
Examiner Tips and Tricks
Any multiple of that equation, for example −6x − y + 103 = 0 , would also be a correct
answer.
(5 marks)
for more: tyrionpapers.com
PREDICTED PAPER
AQA A-level
MATHEMATICS
7357/1
Paper 1
Mark scheme
,1 (a) A curve C has parametric equations
x = 2t 3 y = 4t − 1 t >0
dy
Find an expression for in terms of t .
dx
Answer
Examiner Tips and Tricks
⎛⎜ dy ⎞⎟
⎜ ⎟
dy ⎝ dt ⎠
For parametric equations, =
dx ⎛⎜ dx ⎞⎟
⎜ ⎟
⎝ dt ⎠
Differentiate with respect to
Differentiate with respect to
[M1]
Divide
Examiner Tips and Tricks
dy
Make sure you divide in the correct order, with on the top!
dt
for more: tyrionpapers.com
, That fraction can be simplified
[M1 A1]
Examiner Tips and Tricks
dy 4
= 2 would also be a correct answer here, or you could also use laws of indices to
dx 6t
dy 2 −2
rewrite the answer as, for example, = t .
dx 3
(3 marks)
(b) Find the equation of the normal to C at the point ( 16, 7) .
Give your answer in the form ax + by + c = 0 where a , b and c are integers to be found.
Answer
Examiner Tips and Tricks
dy
The equation for is in terms of t , so in order to find the gradient of the tangent line
dx
you need to find the t value at the point ( 16, 7) .
At the point , , so
[M1]
Examiner Tips and Tricks
for more: tyrionpapers.com
, You can check your t value by substituting it into the x equation, to make sure it gives 16:
x = 2(2) 3 = 2(8) = 16 ✓
Find the gradient of the tangent by substituting into the formula for
[M1]
The gradient of the normal is the negative reciprocal of that
[A1]
Use the equation of a line , with and
Rearrange into the required form
[A1]
Examiner Tips and Tricks
Any multiple of that equation, for example −6x − y + 103 = 0 , would also be a correct
answer.
(5 marks)
for more: tyrionpapers.com
