Standard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate implicitly, substitute a point to find the gradient, then find the normal equation. While it involves multiple steps (differentiate, solve for dy/dx, evaluate at a point, find perpendicular gradient, write equation), each step is routine and follows a standard algorithm taught in P4. The algebraic manipulation is moderate but not challenging. This is slightly easier than average since it's a textbook application of implicit differentiation without any conceptual surprises.
The curve \(C\) is defined by the equation
$$8x^3 - 3y^2 + 2xy = 9$$
Find an equation of the normal to \(C\) at the point \((2, 5)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.
[7]
8 x 3 − 3 y 2 + 2 x y = 9 → 2 4 x 2 − 6 y + 2 y + 2 x = 0
Answer
Marks
d x d x
A1
d y d y d y 5 3
( 2 , 5 ) : 9 6 − 3 0 + 1 0 + 4 = 0 = ... =
Answer
Marks
d x d x d x 1 3
M1
dy 53 d x 1 3
= or = o.e.
Answer
Marks
dx 13 d y 5 3
A1
1 3 ( )
Normal: y − 5 = " − " x − 2
Answer
Marks
5 3
M1
13x+53y−291=0
A1
(7)
(7 marks)
dy
M1: Differentiates one term in y to a correct form. Award for either −3y2 →ky or
dx
dy
2xy→y+x Accept equivalents if they differentiate with respect to y instead.
dx
dM1: Differentiates both y terms to a correct form
Award for Both − 3 y 2 → k y
d
d
y
x
and 2 x y y x
d
d
y
x
→ +
A1: Fully correct differentiation. (o.e. if with respect to y). Allow y' for
d
d
y
x
You may see 24x2dx−6ydy+2ydx+2xdy=0which is fine.
Note that
d
d
y
x
= 2 4 x 2 − 6 y
d
d
y
x
+ 2 y + 2 x
d
d
y
x
= 0 is A0 unless the extra
d
d
y
x
= is subsequently omitted
M1: For
• either rearranging to find
d
d
y
x
(or
d
d
x
y
) in terms of both x and y and then substituting both (2, 5)
• or substituting both (2, 5) then rearranging to find
d
d
y
x
(or
d
d
x
y
)
FYI
d
d
y
x
=
2 4
6
x
y
2
−
+
2
2
x
y
o.e.
It is dependent upon having exactly two terms in
d
d
y
x
, one each coming from differentiating the 3y2
and the 2 x y .
A1: Correct value for either
d
d
y
x
=
5
1
3
3
(accept awrt 4.08) or
d
d
x
y
=
1
5
3
3
(awrt 0.245)
M1: Correct method for the equation of the normal at (2, 5). E.g. y − 5 = " −
1
5
3
3
"
(
x − 2
)
It is dependent upon them having attempted differentiation (may be only one term in
d
d
y
x
) and using
x=2,y=5 to find the value of −
d
d
x
y
PMT
dy
which must be the negative reciprocal of their
dx
If the form y =mx+c is used then the method must proceed as far as c = ...
A1: Correct answer or any integer multiple of this. The = 0 must be seen
Answer
Marks
Guidance
Question
Scheme
Marks
Question 3:
3 | 8 x 3 − 3 y 2 + 2 x y = 9
d y d y
Either 3 y 2 → k y or 2 x y y x → +
d x d x | M1
dy dy
Both 3y2 →ky and 2xy→y+x
dx dx | dM1
d y d y
8 x 3 − 3 y 2 + 2 x y = 9 → 2 4 x 2 − 6 y + 2 y + 2 x = 0
d x d x | A1
d y d y d y 5 3
( 2 , 5 ) : 9 6 − 3 0 + 1 0 + 4 = 0 = ... =
d x d x d x 1 3 | M1
dy 53 d x 1 3
= or = o.e.
dx 13 d y 5 3 | A1
1 3 ( )
Normal: y − 5 = " − " x − 2
5 3 | M1
13x+53y−291=0 | A1
(7)
(7 marks)
dy
M1: Differentiates one term in y to a correct form. Award for either −3y2 →ky or
dx
dy
2xy→y+x Accept equivalents if they differentiate with respect to y instead.
dx
dM1: Differentiates both y terms to a correct form
Award for Both − 3 y 2 → k y
d
d
y
x
and 2 x y y x
d
d
y
x
→ +
A1: Fully correct differentiation. (o.e. if with respect to y). Allow y' for
d
d
y
x
You may see 24x2dx−6ydy+2ydx+2xdy=0which is fine.
Note that
d
d
y
x
= 2 4 x 2 − 6 y
d
d
y
x
+ 2 y + 2 x
d
d
y
x
= 0 is A0 unless the extra
d
d
y
x
= is subsequently omitted
M1: For
• either rearranging to find
d
d
y
x
(or
d
d
x
y
) in terms of both x and y and then substituting both (2, 5)
• or substituting both (2, 5) then rearranging to find
d
d
y
x
(or
d
d
x
y
)
FYI
d
d
y
x
=
2 4
6
x
y
2
−
+
2
2
x
y
o.e.
It is dependent upon having exactly two terms in
d
d
y
x
, one each coming from differentiating the 3y2
and the 2 x y .
A1: Correct value for either
d
d
y
x
=
5
1
3
3
(accept awrt 4.08) or
d
d
x
y
=
1
5
3
3
(awrt 0.245)
M1: Correct method for the equation of the normal at (2, 5). E.g. y − 5 = " −
1
5
3
3
"
(
x − 2
)
It is dependent upon them having attempted differentiation (may be only one term in
d
d
y
x
) and using
x=2,y=5 to find the value of −
d
d
x
y
PMT
dy
which must be the negative reciprocal of their
dx
If the form y =mx+c is used then the method must proceed as far as c = ...
A1: Correct answer or any integer multiple of this. The = 0 must be seen
Question | Scheme | Marks
The curve $C$ is defined by the equation
$$8x^3 - 3y^2 + 2xy = 9$$
Find an equation of the normal to $C$ at the point $(2, 5)$, giving your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers.
[7]
\hfill \mbox{\textit{Edexcel P4 2024 Q3 [7]}}