AS June 2024 Q5
5. The parabola \(C\) has equation \(y^2 = 16x\)
The point \(P\) on \(C\) has \(y\) coordinate \(p\), where \(p\) is a positive constant.
The line \(l\) is the reflection of the tangent to \(C\) at \(P\) in the directrix of \(C\).
Given that \(l\) passes through the focus of \(C\),
| Scheme | Marks | AO |
|---|---|---|
| At \(P\), \(x = \dfrac{p^2}{16}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8}{p}\) so \(y - p = \dfrac{8}{p}\left(x - \dfrac{p^2}{16}\right)\) or At \(P\), \(x = \dfrac{p^2}{16}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8}{p}\) so \(y = \dfrac{8}{p}x + c,\ p = \dfrac{8}{p} \times \dfrac{p^2}{16} + c \Rightarrow c = \ldots\) | M1 | 1.1b |
| \(\Rightarrow py - p^2 = 8x - \dfrac{p^2}{2} \Rightarrow 2py = 16x + p^2\ *\) | A1* | 2.1 |
| (2) |
Notes
M1: Uses equation of line formula with \(y_1 = p\), \(x_1 = \dfrac{p^2}{16}\) and \(m = \dfrac{8}{p}\).
Alternatively uses \(y = mx + c\) with \(y = p\), \(x = \dfrac{p^2}{16}\) and \(m = \dfrac{8}{p}\) and finds \(c\) in terms of \(p\).
A1*: Completes correctly to the given equation, no error seen.
| Scheme | Marks | AO |
|---|---|---|
| \(x = -4\) oe (e.g. \(x + 4 = 0\)) | B1 | 2.2a |
| (1) |
Notes
B1: Deduces the correct equation, accept any form and isw once a correct form is seen.
For part (c), there may be several different attempts so mark the best single attempt i.e. do not “mix and match”.
So e.g. if you are awarding B1 for the coordinates of the focus then follow ALT 3 otherwise award M1 for using the correct focus appropriately in the other alternatives.
There may be methods other than those shown.
If you are not sure if a particular response deserves credit then use review.
| Scheme | Marks | AO |
|---|---|---|
| MAIN Gradient of \(l\) is \(-\dfrac{8}{p}\) | B1 | 2.2a |
| Tangent meets directrix when \(2py = 16 \times -4 + p^2 \Rightarrow y = \dfrac{p^2 - 64}{2p}\) | M1 A1 | 3.1a 1.1b |
| Equation of \(l\) is \(y - \dfrac{p^2 - 64}{2p} = \text{“}-\dfrac{8}{p}\text{”}(x + 4)\) or \(y - 0 = \text{“}-\dfrac{8}{p}\text{”}(x - 4)\) | M1 | 3.1a |
| Focus on \(l \Rightarrow 0 - \dfrac{p^2 - 64}{2p} = -\dfrac{8}{p}(4 + 4) \Rightarrow p = \ldots\) or Point on directrix \(\Rightarrow \dfrac{p^2 - 64}{2p} = \text{“}-\dfrac{8}{p}\text{”}(-4 - 4) \Rightarrow p = \ldots\) | ddM1 | 1.1b |
| \(\Rightarrow p^2 - 64 = 128 \Rightarrow p = 8\sqrt{3}\) | A1 | 1.1b |
| (6) | ||
| (9 marks) |
Notes
B1: Correct gradient for \(l\) seen or implied by working.
M1: Uses \(x = -4\) in the tangent to find the \(y\) coordinate of the intersection of tangent and directrix.
A1: Correct \(y\) coordinate.
M1: Attempts the equation of \(l\) using their “changed” gradient and their point of intersection of the tangent with the directrix of the form \((-4, \mathrm{f}(p))\) or using the focus \((4, 0)\) correctly placed in their equation.
ddM1: Uses the focus \((4, 0)\) is on the line or uses their point of intersection of the tangent with the directrix is on the line and solves to find a real non-zero value for \(p\)
Depends on both previous method marks.
A1: Correct value for \(p\), accept if the negative is also included. Accept \(\sqrt{192}\)
(c) Alt 1
| Scheme | Marks | AO |
|---|---|---|
| Gradient of \(l\) is \(-\dfrac{8}{p}\) | B1 | 2.2a |
| Reflection of point \(P\) has \(x\) coordinate \(x' = -4 - \left(4 + \dfrac{p^2}{16}\right) = -8 - \dfrac{p^2}{16}\) | M1 A1 | 3.1a 1.1b |
| Equation of \(l\) is \(y - p = \text{“}-\dfrac{8}{p}\text{”}\left(x - \left(-8 - \dfrac{p^2}{16}\right)\right)\) | M1 | 3.1a |
| Focus on \(l \Rightarrow 0 - p = -\dfrac{8}{p}\left(4 + 8 + \dfrac{p^2}{16}\right) \Rightarrow p = \ldots\) | ddM1 | 1.1b |
| \(\Rightarrow p^2 = 96 + \dfrac{p^2}{2} \Rightarrow p = 8\sqrt{3}\) | A1 | 1.1b |
| (6) |
B1: Correct gradient for \(l\) seen or implied by working.
M1: Attempts to find the \(x\) coordinate of the reflection of the point \(P\) in the directrix.
A1: Correct coordinate.
M1: Attempts the equation of \(l\) using their “changed” gradient and coordinate.
ddM1: Uses the focus \((4, 0)\) is on the line and solves to find a real non-zero value for \(p\)
Depends on both previous method marks.
A1: Correct value for \(p\), accept if the negative is also included. Accept \(\sqrt{192}\)
(c) Alt 2
| Scheme | Marks | AO |
|---|---|---|
| Gradient of \(l\) is \(-\dfrac{8}{p}\) | B1 | 2.2a |
| Tangent meets directrix when \(2py = 16 \times -4 + p^2 \Rightarrow y = \dfrac{p^2 - 64}{2p}\) | M1 A1 | 3.1a 1.1b |
| Also \(y\) coordinate of intersection satisfies \(-\dfrac{8}{p} = \dfrac{y}{-8}\) or \(-\dfrac{8}{p} = \dfrac{p^2 - 64}{-16p}\) | M1 | 3.1a |
| \(\Rightarrow \dfrac{8}{p} = \dfrac{p^2 - 64}{16p} \Rightarrow p = \ldots\) | ddM1 | 1.1b |
| \(\Rightarrow p^2 - 64 = 128 \Rightarrow p = 8\sqrt{3}\) | A1 | 1.1b |
| (6) |
B1: Correct gradient for \(l\) seen or implied by working.
M1: Uses \(x = -4\) in the tangent to find the \(y\) coordinate of the intersection of tangent and directrix.
A1: Correct \(y\) coordinate.
M1: Uses gradient from intersection of tangent and directrix to focus \((4, 0)\) to form equation in \(y\) and \(p\) or \(p\) only if their \(y\) coordinate is already substituted.
ddM1: Substitutes the \(y\) coordinate of the intersection of tangent and directrix and solves to find a real non-zero value for \(p\) Depends on both previous method marks.
A1: Correct value for \(p\), accept if the negative is also included. Accept \(\sqrt{192}\)
(c) Alt 3
| Scheme | Marks | AO |
|---|---|---|
| Focus of \(C\) is \((4, 0)\) | B1 | 2.2a |
| Reflection of focus in directrix is \((4 - 2(4 - -4), 0) = (-12, 0)\) | M1 A1 | 3.1a 1.1b |
| So tangent passes through \((-12, 0) \Rightarrow 0 = 16 \times -12 + p^2\) | M1 | 3.1a |
| \(\Rightarrow p = \ldots\) | ddM1 | 1.1b |
| \(\Rightarrow p = 8\sqrt{3}\) | A1 | 1.1b |
| (6) |
B1: Correct focus for \(C\) \((4, 0)\) stated or implied by working or e.g. seen on a sketch.
M1: Attempts to find the reflection of the focus of \(C\) in the directrix.
A1: Correct reflected focus point \((-12, 0)\)
M1: Realises if \(l\) passes through focus then the tangent must pass through the reflection of the focus and substitutes their \(x\) from reflection and \(y = 0\) into the tangent equation.
ddM1: Solves to find a real non-zero value for \(p\)
Depends on both previous method marks.
A1: Correct value for \(p\), accept if the negative is also included. Accept \(\sqrt{192}\)
(c) ALT 4
| Scheme | Marks | AO |
|---|---|---|
| Gradient of \(l\) is \(-\dfrac{8}{p}\) | B1 | 2.2a |
| Equation of \(l\) is \(y - 0 = -\dfrac{8}{p}(x - 4)\) | M1 A1 | 3.1a 1.1b |
| Lines intersect when \(\text{“}-\dfrac{8}{p}\text{”}(x - 4) = \dfrac{8x}{p} + \dfrac{p}{2}\) | M1 | 3.1a |
| Lines intersect when \(x = -4 \Rightarrow \text{“}-\dfrac{8}{p}\text{”}(-4 - 4) = \dfrac{8(-4)}{p} + \dfrac{p}{2} \Rightarrow p = \ldots\) | ddM1 | 1.1b |
| \(\Rightarrow p^2 - 64 = 128 \Rightarrow p = 8\sqrt{3}\) | A1 | 1.1b |
| (6) |
B1: Correct gradient for \(l\) seen or implied by working.
M1: Uses the focus \((4, 0)\) with their gradient to form an equation for \(l\)
A1: Correct equation.
M1: Solves simultaneously with the equation of the tangent to obtain an equation in \(x\) and \(p\)
ddM1: Substitutes \(x = -4\) and solves to find a real non-zero value for \(p\)
Depends on both previous method marks.
A1: Correct value for \(p\), accept if the negative is also included. Accept \(\sqrt{192}\)