# Question 5 (Parabola):

## Part (a):

| Working | Mark | Guidance |
|---------|------|----------|
| At $P$, $x=\frac{p^2}{16}$ and $\frac{dy}{dx}=\frac{8}{p}$ so $y-p=\frac{8}{p}\left(x-\frac{p^2}{16}\right)$ | M1 | Uses equation of line formula with $y_1=p$, $x_1=\frac{p^2}{16}$ and $m=\frac{8}{p}$. Alternatively uses $y=mx+c$ and finds $c$ in terms of $p$ |
| $\Rightarrow py - p^2 = 8x - \frac{p^2}{2} \Rightarrow 2py = 16x + p^2$ | A1* | Completes correctly to given equation, no error seen |

## Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| $x = -4$ oe (e.g. $x+4=0$) | B1 | Deduces correct equation, accept any form once correct form is seen |

## Part (c) MAIN:

| Working | Mark | Guidance |
|---------|------|----------|
| Gradient of $l$ is $-\frac{8}{p}$ | B1 | Correct gradient for $l$ seen or implied by working |
| Tangent meets directrix when $2py = 16\times-4+p^2 \Rightarrow y = \frac{p^2-64}{2p}$ | M1, A1 | Uses $x=-4$ in tangent to find $y$ coordinate of intersection; correct $y$ coordinate |
| Equation of $l$ is $y - \frac{p^2-64}{2p} = \text{"}-\frac{8}{p}\text{"}(x+4)$ or $y-0 = \text{"}-\frac{8}{p}\text{"}(x-4)$ | M1 | Attempts equation of $l$ using "changed" gradient and point of intersection $(-4, f(p))$ or using focus $(4,0)$ |
| Focus on $l \Rightarrow 0 - \frac{p^2-64}{2p} = -\frac{8}{p}(4+4) \Rightarrow p = \ldots$ | ddM1 | Uses focus $(4,0)$ on line or uses point of intersection of tangent with directrix on line, solves for real non-zero $p$. **Depends on both previous M marks** |
| $\Rightarrow p^2 - 64 = 128 \Rightarrow p = 8\sqrt{3}$ | A1 | Correct value for $p$; accept if negative also included. Accept $\sqrt{192}$ |

## Part (c) Alt 1:

| Working | Mark | Guidance |
|---------|------|----------|
| Gradient of $l$ is $-\frac{8}{p}$ | B1 | Correct gradient seen or implied |
| Reflection of $P$ has $x$ coordinate $x' = -4-\left(4+\frac{p^2}{16}\right) = -8-\frac{p^2}{16}$ | M1, A1 | Attempts $x$ coordinate of reflection of $P$ in directrix; correct coordinate |
| Equation of $l$ is $y-p = \text{"}-\frac{8}{p}\text{"}\left(x-\left(-8-\frac{p^2}{16}\right)\right)$ | M1 | Attempts equation of $l$ using changed gradient and coordinate |
| Focus on $l \Rightarrow 0-p = -\frac{8}{p}\left(4+8+\frac{p^2}{16}\right) \Rightarrow p=\ldots$ | ddM1 | Uses focus $(4,0)$ on line, solves for real non-zero $p$. Depends on both M marks |
| $\Rightarrow p^2 = 96 + \frac{p^2}{2} \Rightarrow p = 8\sqrt{3}$ | A1 | Correct value; accept $\sqrt{192}$ |

## Part (c) Alt 2:

| Working | Mark | Guidance |
|---------|------|----------|
| Gradient of $l$ is $-\frac{8}{p}$ | B1 | Correct gradient seen or implied |
| $2py = 16\times-4+p^2 \Rightarrow y = \frac{p^2-64}{2p}$ | M1, A1 | Uses $x=-4$ in tangent; correct $y$ coordinate |
| $-\frac{8}{p} = \frac{y}{-8}$ or $-\frac{8}{p} = \frac{p^2-64}{-16p}$ | M1 | Uses gradient from intersection of tangent and directrix to focus $(4,0)$ |
| $\frac{8}{p} = \frac{p^2-64}{16p} \Rightarrow p = \ldots$ | ddM1 | Substitutes $y$ coordinate and solves for real non-zero $p$. Depends on both M marks |
| $p^2 - 64 = 128 \Rightarrow p = 8\sqrt{3}$ | A1 | Correct value; accept $\sqrt{192}$ |

## Question (c) Alt 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Focus of $C$ is $(4, 0)$ | **B1** | Correct focus for $C(4,0)$ stated or implied by working or e.g. seen on a sketch |
| Reflection of focus in directrix is $\left(4-2(4--4),0\right)=(-12,0)$ | **M1 A1** | M1: Attempts to find the reflection of the focus of $C$ in the directrix. A1: Correct reflected focus point $(-12, 0)$ |
| So tangent passes through $(-12,0) \Rightarrow 0 = 16 \times -12 + p^2$ | **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 |
| $\Rightarrow p = \ldots$ | **ddM1** | Solves to find a real non-zero value for $p$. **Depends on both previous method marks** |
| $\Rightarrow p = 8\sqrt{3}$ | **A1** | Correct value for $p$; accept if the negative is also included. Accept $\sqrt{192}$ |

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## Question (c) Alt 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of $l$ is $-\dfrac{8}{p}$ | **B1** | Correct gradient for $l$ seen or implied by working |
| Equation of $l$ is $y - 0 = -\dfrac{8}{p}(x-4)$ | **M1 A1** | M1: Uses the focus $(4,0)$ with their gradient to form an equation for $l$. A1: Correct equation |
| Lines intersect when $-\dfrac{8}{p}(x-4) = \dfrac{8x}{p} + \dfrac{p}{2}$ | **M1** | Solves simultaneously with the equation of the tangent to obtain an equation in $x$ and $p$ |
| Lines intersect when $x=-4 \Rightarrow -\dfrac{8}{p}(-4-4) = \dfrac{8(-4)}{p} + \dfrac{p}{2} \Rightarrow p = \ldots$ | **ddM1** | Substitutes $x=-4$ and solves to find a real non-zero value for $p$. **Depends on both previous method marks** |
| $\Rightarrow p^2 - 64 = 128 \Rightarrow p = 8\sqrt{3}$ | **A1** | Correct value for $p$; accept if the negative is also included. Accept $\sqrt{192}$ |