Mathematics · May / June 2025 · Paper 41 · Question 5
YMA01
As printed in the paper
The question as text
5. The curve C has parametric equations t 3 2 2 x y t t 1 1 t 1 The point P, where t = 2, lies on C.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[5]
−
−−
+
and d
y
d
x
2(1
t
)
( 1)(3
2 )
t
5
t = −
=
=
2
t
d
t
(1
−
t
)
2
(1
−
t
)
2
d
d
y
( 1)
−
2
4
(
)
At
t =
2
=
×−
2(2)
= −
d
x
5
5
−
−
( 7,
3)
5
+
=
+
y
3
(
x
7)
4
−
+
=
5
x
4
y
23
0
5(b)[4]
3
+
2
t
x
−
3
=
→=
x
t
1
−
t
x
+
2
−
2
+
2
−
−
2
−
x
3
(
x
2)
(
x
3)
10
x
5
y
= −
1
=
=
(
)
(
)
x
+
2
+
2
+
2
x
2
x
2
x ≠−
2
3
+
2
t
5
=
=
−
FYI
x
2
−
−
1
t
1
t
(a) Note: A cartesian approach in (a) can score a maximum of 1 mark (the B1) as it does not satisfy
the demand of the question.
Attempts to find d
y
by finding d
y
d
x
M1:
x at
t =
2
÷
.
d
d
t
d
t
Condone slips when differentiating d
x
t and d
y
t but look for a minimum of
d
d
d
x
α
(1
−
t
)
±
β
(3
+
2 )
t
=
α
>
with
0
or equivalent using the product rule or division and chain rule
d
t
(1
−
t
)
2
d
y
t = ±
t
d
into d
y
d
x
Substitution oft =
2
÷
which may have been simplified incorrectly. Note that a
d
t
d
t
(
)
−
2 1
t
−
t
2
d
y
simplified
=
d
x
5
uv
'
−
vu
'
Note that using the quotient rule the wrong way around
would be M0
v
2
4
following M1. It may be implied by a normal gradient of 5
A1:
−
4
5
−
−
B1:
( 7,
3)
which may be implied from terms within an attempt at the equation of the normal
Attempts to find the equation of the normal at P. They must be using the negative reciprocal of their d
M1:
with their point P . It is dependent on having found d
y
d
y
t =
2
x from
÷
their
their
d
d
t
attempts at d
x
t and d
y
t or what they believe are d
x
t and d
y
t so 0, 0, 1, 1, 0 is possible. Condone one sign
d
d
d
d
slip when using their coordinate of P. e.g( 7,
−
−
3)
→
(7,
−
3)
=
+
If the form y
mx
c
is used the method must proceed as far as c = …
A1:
5
x
−
4
y
+
23
=
0
or any integer multiple with terms on one side of an equation. The order of the terms is not
important. ISW after sight of a correct answer in a correct form
……………………………………………………………………………………………………….
Other more complicated methods do exist so look carefully at what is attempted. For example, candidates could
find d
x
by finding d
x
d
y
d
x
t =
÷
−
y at
2
and use the value of
as the gradient of the normal.
d
d
t
d
t
d
y
…………………………………………………………………………………………………………
(b)
M1:
Attempts to rearrange the parametric equation for x to make t the subject. Expect to see cross multiplying
f ( )
x
(or equivalent) and collection of terms with
t
=
where f(x) and g(x) are linear functions in x.
g( )
x
dM1: Substitutes their expression for t in terms for x into the parametric equation for y and attempts to
manipulate the righthand side into a single fraction. It is dependent on the previous method mark.
2
f( )
x
(g( ))
x
2
−
(f( ))
x
2
= −
=
For example, look for
y
1
(
)
g( )
x
2
g( )
x
−
10
x
5
A1:
y
=
(
)
+
2
x
2
k = − but must follow correct work. E.g. Sight of (
)
x ≠− or
x +
B1:
2
2
2
o.e. on the denominator of either the
expression for t or the expression for y.
…………………………………………………………………………………………………………………………………………
Other more complicated methods do exist but are unlikely to lead to any success. If you see a solution that you feel
deserves credit please send to review and alert your TL.
3
+
2 1
−
y
=
−
=
For example, if they start with
t
1
y
and then write down
x
the first M mark will not be scored
1
−
1
−
y
until they make 1
−
y
the subject (in terms of just x) which reverts to the original method
(
)
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