# Question 16:

## Part (a):

| Working | Mark | Guidance |
|---------|------|----------|
| $\dfrac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} = \dfrac{4\sec^2 t \tan t}{2\sec^2 t} (= 2\tan t)$ | M1, A1 | Must use parametric differentiation; attempt to differentiate both parameters and divide correctly |
| At $t = \dfrac{\pi}{4}$: $\dfrac{dy}{dx} = 2$, $x = 3$, $y = 7$ | M1 | Getting at least two correct |
| Attempts normal equation: $y - 7 = -\dfrac{1}{2}(x-3)$ | M1 | Correct attempt at normal using negative reciprocal of their $\frac{dy}{dx}$ |
| $y = -\dfrac{1}{2}x + \dfrac{17}{2}$ * | A1* | Proceeds to given answer with no errors |

## Part (b):

| Working | Mark | Guidance |
|---------|------|----------|
| Use $\sec^2 t = 1 + \tan^2 t \Rightarrow \dfrac{y-3}{2} = 1 + \left(\dfrac{x-1}{2}\right)^2$ | M1 | |
| $y - 3 = 2 + \dfrac{(x-1)^2}{2} \Rightarrow y = \dfrac{1}{2}(x-1)^2 + 5$ * | A1* | |

## Part (c):

| Working | Mark | Guidance |
|---------|------|----------|
| **Lower limit for $k$:** $\frac{1}{2}(x-1)^2 + 5 = -\frac{1}{2}x + k \Rightarrow x^2 - x + (11-2k) = 0$ | M1 | |
| $b^2 - 4ac = 1 - 4(11-2k) = 0 \Rightarrow k = \ldots$ | | |
| $k = \dfrac{43}{8}$ | A1 | |
| **Upper limit for $k$:** $(x,y)_{t=-\pi/4}: x = -1,\ y = 7$ | M1 | |
| $(-1, 7):\ y = -\dfrac{1}{2}x + k \Rightarrow 7 = \dfrac{1}{2} + k \Rightarrow k = \ldots$ | | |
| $k = \dfrac{13}{2}$ | A1 | |
| $\dfrac{43}{8} < k \leqslant \dfrac{13}{2}$ | A1 | |

# Question (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to use $\sec^2 t = 1 + \tan^2 t$ to obtain an equation involving $y$ and $(x-1)^2$ | M1 | e.g. $y = 2\sec^2 t + 3 = 2(1+\tan^2 t)+3 = 2\left(1+\left(\frac{x-1}{2}\right)^2\right)+3$ |
| $y = \frac{1}{2}(x-1)^2 + 5$ | A1* | Proceeds with clear argument to given answer with no errors |

**Alternative 1:** Uses given result, substitutes for $x$, attempts $\sec^2 t = 1+\tan^2 t$; proceeds to $y$ parameter with minimal conclusion e.g. "$= y$", QED | M1, A1 |

**Alternative 2:** Uses $x$ parameter to obtain $t$ in terms of arctan, substitutes into $y$, uses $\sec^2 t = 1+\tan^2 t$ | M1, A1 |

**Alternative 3:** Uses $\frac{dy}{dx}$ from part (a), integrates, uses $(3,7)$ to find $c$ | M1, A1 | Allow marks for (b) to score anywhere in solution |

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# Question (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Full attempt to find **lower** limit for $k$: sets $\frac{1}{2}(x-1)^2+5 = -\frac{1}{2}x+k$, rearranges to 3TQ, attempts $b^2-4ac$ | M1 | $x^2 - x+(11-2k)=0 \Rightarrow b^2-4ac = 1-4(11-2k)=0$ |
| $k = \frac{43}{8}$ | A1 | Value may appear in inequality e.g. $k>\frac{43}{8}$, $k<\frac{43}{8}$ |
| **Calculus alternative for lower limit:** $\frac{dy}{dx}=x-1=-\frac{1}{2} \Rightarrow x=\frac{1}{2}$; $y=\frac{1}{2}\left(\frac{1}{2}-1\right)^2+5=\frac{41}{8}$; $\frac{41}{8}=-\frac{1}{4}+k \Rightarrow k=\frac{43}{8}$ | M1, A1 | Score M1 for $\frac{dy}{dx}$ = linear expression in $x$, sets $=-\frac{1}{2}$, solves for $x$, substitutes to find $y$, uses $y=-\frac{1}{2}x+k$ |
| **Parameters alternative for lower limit:** substitutes parametric forms into $y=-\frac{1}{2}x+k$, uses $\sec^2 t=1+\tan^2 t$, rearranges to 3TQ, sets $b^2-4ac=0 \Rightarrow k=\frac{43}{8}$ | M1, A1 | |
| Full attempt to find **upper** limit: uses $t=-\frac{\pi}{4}$, finds $x$ and $y$, substitutes into $y=-\frac{1}{2}x+k$, solves for $k$ | M1 | |
| $k = \frac{13}{2}$ | A1 | Value may appear in inequality |
| $\frac{43}{8} < k \leqslant \frac{13}{2}$ | A1 | Allow e.g. $\left(k\leqslant\frac{13}{2}\ \text{and}\ k>\frac{43}{8}\right)$, $\left(\frac{43}{8},\frac{13}{2}\right]$; do **not** allow $\cup$, or $\leqslant\frac{13}{2}\ \text{or}\ >\frac{43}{8}$, or in terms of $x$ |