| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola tangent intersection problems |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring parametric differentiation, tangent equations, perpendicular lines, and knowledge of parabola focus/directrix properties. While systematic, it demands multiple techniques and geometric insight about conic sections beyond standard A-level, placing it moderately above average difficulty. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{dy}{dx} = a^{\frac{1}{2}}x^{-\frac{1}{2}}\) or \(2y\dfrac{dy}{dx} = 4a\) | M1 | \(\dfrac{dy}{dx} = \dfrac{2a}{2aq}\) OK |
| Gradient of tangent is \(\dfrac{1}{q}\) | A1 | |
| Equation of tangent: \(y - 2aq = \dfrac{1}{q}(x - aq^2)\) | M1 | Use of \(y = mx + c\) to find \(c\) OK |
| So \(yq = x + aq^2\) | A1 (4) | Correct solution only for final A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R\) has coordinates \((0,\ aq)\) | B1 | |
| Line \(l\) has equation \(y - aq = -qx\) | M1 A1 (3) | \(-1/\)(their gradient from part a) in equation OK for M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(y = 0\), \(x = a\), so line \(l\) passes through \((a,\ 0)\), the focus | B1 (1) | Must attempt \(y=0\) or \(x=a\) to show correct coordinates of \(R\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Line \(l\) meets directrix when \(x = -a\): then \(y = 2aq\), so coordinates are \((-a,\ 2aq)\) | M1 A1 (2) [10] | Substitute \(x = -a\) for M1; both coordinates correct for A1 |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = a^{\frac{1}{2}}x^{-\frac{1}{2}}$ or $2y\dfrac{dy}{dx} = 4a$ | M1 | $\dfrac{dy}{dx} = \dfrac{2a}{2aq}$ OK |
| Gradient of tangent is $\dfrac{1}{q}$ | A1 | |
| Equation of tangent: $y - 2aq = \dfrac{1}{q}(x - aq^2)$ | M1 | Use of $y = mx + c$ to find $c$ OK |
| So $yq = x + aq^2$ | A1 (4) | Correct solution only for final A1 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R$ has coordinates $(0,\ aq)$ | B1 | |
| Line $l$ has equation $y - aq = -qx$ | M1 A1 (3) | $-1/$(their gradient from part a) in equation OK for M1 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $y = 0$, $x = a$, so line $l$ passes through $(a,\ 0)$, the focus | B1 (1) | Must attempt $y=0$ or $x=a$ to show correct coordinates of $R$ |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Line $l$ meets directrix when $x = -a$: then $y = 2aq$, so coordinates are $(-a,\ 2aq)$ | M1 A1 (2) [10] | Substitute $x = -a$ for M1; both coordinates correct for A1 |
---8. A parabola has equation $y ^ { 2 } = 4 a x , a > 0$. The point $Q \left( a q ^ { 2 } , 2 a q \right)$ lies on the parabola.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the tangent to the parabola at $Q$ is
$$y q = x + a q ^ { 2 } .$$
This tangent meets the $y$-axis at the point $R$.
\item Find an equation of the line $l$ which passes through $R$ and is perpendicular to the tangent at $Q$.
\item Show that $l$ passes through the focus of the parabola.
\item Find the coordinates of the point where $l$ meets the directrix of the parabola.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2009 Q8 [10]}}