Question 5 - A-Level Maths

OCR PURE — Question 5 5 marks

Exam Board	OCR
Module	PURE
Marks	5
Paper	Download PDF ↗
Topic	Simultaneous equations
Type	Tangency condition for line and curve
Difficulty	Standard +0.3 This is a standard tangency condition problem requiring substitution of the line equation into the curve equation and setting the discriminant to zero. While it involves multiple algebraic steps (rearranging to quadratic form, applying b²-4ac=0, solving for k), the method is routine and commonly practiced. Slightly above average difficulty due to the algebraic manipulation required, but well within typical A-level expectations.
Spec	1.02c Simultaneous equations: two variables by elimination and substitution 1.07m Tangents and normals: gradient and equations

5 In this question you must show detailed reasoning.
The line \(x + 5 y = k\) is a tangent to the curve \(x ^ { 2 } - 4 y = 10\). Find the value of the constant \(k\).

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Question 5:

Answer	Marks	Guidance
Answer	Marks	Guidance
\((k-5y)^2 - 4y = 10\)	M1*	Substitute for \(x/y\) to eliminate one of the variables. If \(y\) eliminated: \(5x^2+4x-4k-50\ (=0)\)
\(25y^2 + (-4-10k)y + (k^2-10)(=0)\)	A1	Obtain correct (unsimplified) quadratic
Tangent \(\Rightarrow b^2 - 4ac = 0\)	Dep*M1	Uses \(b^2-4ac\) correctly for their quadratic
\((-4-10k)^2 - 4(25)(k^2-10) = 0\)	A1	Fully correct substitution must equal 0. \(16-4(5)(-4k-50)=0\)
\(k = -\frac{127}{10}\) \((-12.7)\)	A1 [5]	\(k\) correct with sufficient working
OR Gradient of line \(= -\frac{1}{5}\)	B1
\(\frac{dy}{dx} = \frac{1}{2}x\)	B1	Correct differentiation. \(2x - 4\frac{dy}{dx}=0\)
\(\frac{1}{2}x = -\frac{1}{5}\)	M1	Equates their derivative with their gradient of line
\(x = -\frac{2}{5}\)	A1	\(x\) from correct working only
\(y = -\frac{123}{50}\ (-2.46) \Rightarrow k = -\frac{127}{10}\ (-12.7)\)	A1	\(k\) from correct working only

# Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(k-5y)^2 - 4y = 10$ | M1* | Substitute for $x/y$ to eliminate one of the variables. If $y$ eliminated: $5x^2+4x-4k-50\ (=0)$ |
| $25y^2 + (-4-10k)y + (k^2-10)(=0)$ | A1 | Obtain correct (unsimplified) quadratic |
| Tangent $\Rightarrow b^2 - 4ac = 0$ | Dep*M1 | Uses $b^2-4ac$ correctly for their quadratic |
| $(-4-10k)^2 - 4(25)(k^2-10) = 0$ | A1 | Fully correct substitution must equal 0. $16-4(5)(-4k-50)=0$ |
| $k = -\frac{127}{10}$ $(-12.7)$ | A1 [5] | $k$ correct with sufficient working |
| **OR** Gradient of line $= -\frac{1}{5}$ | B1 | |
| $\frac{dy}{dx} = \frac{1}{2}x$ | B1 | Correct differentiation. $2x - 4\frac{dy}{dx}=0$ |
| $\frac{1}{2}x = -\frac{1}{5}$ | M1 | Equates their derivative with their gradient of line |
| $x = -\frac{2}{5}$ | A1 | $x$ from correct working only |
| $y = -\frac{123}{50}\ (-2.46) \Rightarrow k = -\frac{127}{10}\ (-12.7)$ | A1 | $k$ from correct working only |

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5 In this question you must show detailed reasoning.\\
The line $x + 5 y = k$ is a tangent to the curve $x ^ { 2 } - 4 y = 10$. Find the value of the constant $k$.

\hfill \mbox{\textit{OCR PURE  Q5 [5]}}

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