Question 7 - A-Level Maths

CAIE FP1 2015 June — Question 7 10 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2015
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Implicit equations and differentiation
Type	Find stationary points
Difficulty	Standard +0.8 This question requires implicit differentiation to find dy/dx, setting it to zero for horizontal tangents, then solving the resulting system of equations (linear constraint with quadratic curve). The nature determination adds complexity. While the techniques are standard for Further Maths, the multi-step reasoning and algebraic manipulation elevate this above routine exercises.
Spec	1.07s Parametric and implicit differentiation

7 The curve \(C\) has equation \(x ^ { 2 } + 2 x y - 4 y ^ { 2 } + 20 = 0\). Show that if the tangent to \(C\) at the point \(( x , y )\) is parallel to the \(x\)-axis then \(x + y = 0\). Hence find the coordinates of the stationary points on \(C\), and determine their nature.

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Question 7:

Answer	Marks	Guidance
Working/Answer	Marks	Notes
\(2x + 2\!\left(x\frac{dy}{dx}+y\right) - 8y\frac{dy}{dx} = 0\)	B1B1	1st B1 for 1st and 3rd terms; 2nd B1 for rest including \(=0\)
\(\frac{dy}{dx}(x-4y) = -(x+y) \Rightarrow y = -x\) when \(\frac{dy}{dx}=0\)	B1
Substituting \(y=-x \Rightarrow x^2 - 2x^2 - 4x^2 + 20 = 0 \Rightarrow 5x^2 = 20 \Rightarrow x = \pm 2\)	M1
Coordinates are \((2,-2)\) and \((-2,2)\)	A1
\(\frac{d^2y}{dx^2}(x-4y) + \frac{dy}{dx}\!\left(1-4\frac{dy}{dx}\right) = -1 - \frac{dy}{dx}\)	M1A1	OE
Substituting either \((2,-2)\) or \((-2,2)\) and \(\frac{dy}{dx}=0\)	dM1
At \((2,-2)\): \(y''(2) = -\frac{1}{10} \Rightarrow\) max	A1
At \((-2,2)\): \(y''(2) = \frac{1}{10} \Rightarrow\) min	A1

## Question 7:

| Working/Answer | Marks | Notes |
|---|---|---|
| $2x + 2\!\left(x\frac{dy}{dx}+y\right) - 8y\frac{dy}{dx} = 0$ | B1B1 | 1st B1 for 1st and 3rd terms; 2nd B1 for rest including $=0$ |
| $\frac{dy}{dx}(x-4y) = -(x+y) \Rightarrow y = -x$ when $\frac{dy}{dx}=0$ | B1 | |
| Substituting $y=-x \Rightarrow x^2 - 2x^2 - 4x^2 + 20 = 0 \Rightarrow 5x^2 = 20 \Rightarrow x = \pm 2$ | M1 | |
| Coordinates are $(2,-2)$ and $(-2,2)$ | A1 | |
| $\frac{d^2y}{dx^2}(x-4y) + \frac{dy}{dx}\!\left(1-4\frac{dy}{dx}\right) = -1 - \frac{dy}{dx}$ | M1A1 | OE |
| Substituting either $(2,-2)$ or $(-2,2)$ and $\frac{dy}{dx}=0$ | dM1 | |
| At $(2,-2)$: $y''(2) = -\frac{1}{10} \Rightarrow$ max | A1 | |
| At $(-2,2)$: $y''(2) = \frac{1}{10} \Rightarrow$ min | A1 | |

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7 The curve $C$ has equation $x ^ { 2 } + 2 x y - 4 y ^ { 2 } + 20 = 0$. Show that if the tangent to $C$ at the point $( x , y )$ is parallel to the $x$-axis then $x + y = 0$.

Hence find the coordinates of the stationary points on $C$, and determine their nature.

\hfill \mbox{\textit{CAIE FP1 2015 Q7 [10]}}

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