Question 7 - A-Level Maths

Edexcel FP1 2018 June — Question 7 8 marks

Exam Board	Edexcel
Module	FP1 (Further Pure Mathematics 1)
Year	2018
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Conic sections
Type	Parabola focus and directrix properties
Difficulty	Standard +0.8 This is a multi-part Further Maths question requiring knowledge of parabola focus-directrix properties, coordinate geometry, and perpendicular line conditions. Part (a) involves finding the focus and directrix from standard form, then deriving a line equation (routine but requires careful algebra). Part (b) requires using the parametric form of a parabola and applying perpendicularity conditions, involving more sophisticated problem-solving than typical A-level questions but still following standard FP1 techniques.
Spec	1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 1.03d Circles: equation (x-a)^2+(y-b)^2=r^2

The parabola $C$ has equation $y ^ { 2 } = 4 a x$, where $a$ is a positive constant. The point $S$ is the focus of $C$.

The straight line $l$ passes through the point $S$ and meets the directrix of $C$ at the point $D$.
Given that the $y$ coordinate of $D$ is $\frac { 24 a } { 5 }$,

show that an equation of the line $l$ is $$12 x + 5 y = 12 a$$ The point $P \left( a k ^ { 2 } , 2 a k \right)$, where $k$ is a positive constant, lies on the parabola $C$.
Given that the line segment $S P$ is perpendicular to $l$,
find, in terms of $a$, the coordinates of the point $P$.

Show mark scheme Show mark scheme source

Part (a)

Answer	Marks	Guidance
$m_l = \frac{\frac{24a}{5} - 0}{-a - a} = -\frac{\frac{24a}{5}}{-2a} = -\frac{12}{5}$ or $\frac{y - y_1}{x_2 - x_1} = \frac{x - x_1}{x_2 - x_1}$ or $\frac{y - 0}{0 - \frac{24a}{5}} = \frac{x - a}{-a - a}$	M1	Uses $S(a, 0)$ and $D$ their "$-a$", $\frac{24a}{5}$ to find an expression for the gradient of $l$ or applies the formula $\frac{y - y_1}{x_2 - x_1} = \frac{x - x_1}{x_2 - x_1}$. Can be un-simplified or simplified.
$l: y - 0 = -\frac{12}{5}(x - a) \Rightarrow 5y = -12x + 12a$	A1 *	Correct solution only leading to $12x + 5y = 12a$
$l: 12x + 5y = 12a$ (*)		No errors seen.

ALT (a)

Answer	Marks	Guidance
$y = mx + c$. At $S$, $0 = ma + c$ At $D$, $\frac{24a}{5} = -ma + c$ $\Rightarrow c = \frac{12a}{5}, m = -\frac{12}{5}$	M1	Uses $S(a, 0)$ and $D$ their "$-a$", $\frac{24a}{5}$ to find 2 simultaneous equations and solves to achieve $c = ..., m = ...$
$y = -\frac{12}{5}x + \frac{12a}{5} \Rightarrow 12x + 5y = 12a$ *	A1 *	Correct solution only leading to $12x + 5y = 12a$

Part (b)

Answer	Marks	Guidance
$m_{SP} = \frac{2ak}{ak^2 - a} = \frac{2k}{k^2 - 1}$	M1	Attempts to find the gradient of $SP$
$m_l = -\begin{pmatrix} \frac{ak^2 - a}{2ak} \end{pmatrix}$ or $m_{SP} = -\frac{1}{\left(-\frac{5}{12}\right)} = \frac{12}{5}$	M1	Some evidence of applying $m_l m_2 = -1$
So $\left\{\frac{2k}{k^2 - 1} = \frac{5}{12}\right\} \Rightarrow 24k = 5k^2 - 5$	A1	Correct 3TQ in terms of $k$ in any form.
$\{5k^2 - 24k - 5 = 0\} \Rightarrow (k - 5)(5k + 1) = 0 \Rightarrow k = ...$	M1	Attempt to solve their 3TQ for $k$
$\{As k > 0, so k = 5\} \Rightarrow (25a, 10a)$	M1	Uses their $k$ to find $P$
A1	$(25a, 10a)$

Total: 8 marks

**Part (a)**

| $m_l = \frac{\frac{24a}{5} - 0}{-a - a} = -\frac{\frac{24a}{5}}{-2a} = -\frac{12}{5}$ or $\frac{y - y_1}{x_2 - x_1} = \frac{x - x_1}{x_2 - x_1}$ or $\frac{y - 0}{0 - \frac{24a}{5}} = \frac{x - a}{-a - a}$ | M1 | Uses $S(a, 0)$ and $D$ their "$-a$", $\frac{24a}{5}$ to find an expression for the gradient of $l$ or applies the formula $\frac{y - y_1}{x_2 - x_1} = \frac{x - x_1}{x_2 - x_1}$. Can be un-simplified or simplified. |
| $l: y - 0 = -\frac{12}{5}(x - a) \Rightarrow 5y = -12x + 12a$ | A1 * | Correct solution only leading to $12x + 5y = 12a$ |
| $l: 12x + 5y = 12a$ (*) | | No errors seen. |

**ALT (a)**

| $y = mx + c$. At $S$, $0 = ma + c$ At $D$, $\frac{24a}{5} = -ma + c$ $\Rightarrow c = \frac{12a}{5}, m = -\frac{12}{5}$ | M1 | Uses $S(a, 0)$ and $D$ their "$-a$", $\frac{24a}{5}$ to find 2 simultaneous equations and solves to achieve $c = ..., m = ...$ |
| $y = -\frac{12}{5}x + \frac{12a}{5} \Rightarrow 12x + 5y = 12a$ * | A1 * | Correct solution only leading to $12x + 5y = 12a$ |

**Part (b)**

| $m_{SP} = \frac{2ak}{ak^2 - a} = \frac{2k}{k^2 - 1}$ | M1 | Attempts to find the gradient of $SP$ |
| $m_l = -\begin{pmatrix} \frac{ak^2 - a}{2ak} \end{pmatrix}$ or $m_{SP} = -\frac{1}{\left(-\frac{5}{12}\right)} = \frac{12}{5}$ | M1 | Some evidence of applying $m_l m_2 = -1$ |
| So $\left\{\frac{2k}{k^2 - 1} = \frac{5}{12}\right\} \Rightarrow 24k = 5k^2 - 5$ | A1 | Correct 3TQ in terms of $k$ in any form. |
| $\{5k^2 - 24k - 5 = 0\} \Rightarrow (k - 5)(5k + 1) = 0 \Rightarrow k = ...$ | M1 | Attempt to solve their 3TQ for $k$ |
| $\{As k > 0, so k = 5\} \Rightarrow (25a, 10a)$ | M1 | Uses their $k$ to find $P$ |
| | A1 | $(25a, 10a)$ |

**Total: 8 marks**

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Show LaTeX source

\begin{enumerate}
  \item The parabola $C$ has equation $y ^ { 2 } = 4 a x$, where $a$ is a positive constant. The point $S$ is the focus of $C$.
\end{enumerate}

The straight line $l$ passes through the point $S$ and meets the directrix of $C$ at the point $D$.\\
Given that the $y$ coordinate of $D$ is $\frac { 24 a } { 5 }$,\\
(a) show that an equation of the line $l$ is

$$12 x + 5 y = 12 a$$

The point $P \left( a k ^ { 2 } , 2 a k \right)$, where $k$ is a positive constant, lies on the parabola $C$.\\
Given that the line segment $S P$ is perpendicular to $l$,\\
(b) find, in terms of $a$, the coordinates of the point $P$.\\

\hfill \mbox{\textit{Edexcel FP1 2018 Q7 [8]}}

This paper (9 questions)

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Q1 6 Q2 10 Q3 9 Q4 9 Q5 9 Q6 6 Q7 8 Q8 6 Q9 12

Answer	Marks	Guidance
\(m_{SP} = \frac{2ak}{ak^2 - a} = \frac{2k}{k^2 - 1}\)	M1	Attempts to find the gradient of \(SP\)
\(m_l = -\begin{pmatrix} \frac{ak^2 - a}{2ak} \end{pmatrix}\) or \(m_{SP} = -\frac{1}{\left(-\frac{5}{12}\right)} = \frac{12}{5}\)	M1	Some evidence of applying \(m_l m_2 = -1\)
So \(\left\{\frac{2k}{k^2 - 1} = \frac{5}{12}\right\} \Rightarrow 24k = 5k^2 - 5\)	A1	Correct 3TQ in terms of \(k\) in any form.
\(\{5k^2 - 24k - 5 = 0\} \Rightarrow (k - 5)(5k + 1) = 0 \Rightarrow k = ...\)	M1	Attempt to solve their 3TQ for \(k\)
\(\{As k > 0, so k = 5\} \Rightarrow (25a, 10a)\)	M1	Uses their \(k\) to find \(P\)
A1	\((25a, 10a)\)

Answer	Marks	Guidance
\(m_l = \frac{\frac{24a}{5} - 0}{-a - a} = -\frac{\frac{24a}{5}}{-2a} = -\frac{12}{5}\) or \(\frac{y - y_1}{x_2 - x_1} = \frac{x - x_1}{x_2 - x_1}\) or \(\frac{y - 0}{0 - \frac{24a}{5}} = \frac{x - a}{-a - a}\)	M1	Uses \(S(a, 0)\) and \(D\) their "\(-a\)", \(\frac{24a}{5}\) to find an expression for the gradient of \(l\) or applies the formula \(\frac{y - y_1}{x_2 - x_1} = \frac{x - x_1}{x_2 - x_1}\). Can be un-simplified or simplified.
\(l: y - 0 = -\frac{12}{5}(x - a) \Rightarrow 5y = -12x + 12a\)	A1 *	Correct solution only leading to \(12x + 5y = 12a\)
\(l: 12x + 5y = 12a\) (*)		No errors seen.

Answer	Marks	Guidance
\(y = mx + c\). At \(S\), \(0 = ma + c\) At \(D\), \(\frac{24a}{5} = -ma + c\) \(\Rightarrow c = \frac{12a}{5}, m = -\frac{12}{5}\)	M1	Uses \(S(a, 0)\) and \(D\) their "\(-a\)", \(\frac{24a}{5}\) to find 2 simultaneous equations and solves to achieve \(c = ..., m = ...\)
\(y = -\frac{12}{5}x + \frac{12a}{5} \Rightarrow 12x + 5y = 12a\) *	A1 *	Correct solution only leading to \(12x + 5y = 12a\)