# Question 9:

## Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = \sec t \tan t$ and $\frac{dy}{dt} = \sqrt{3}\sec^2\left(t + \frac{\pi}{3}\right)$ | B1 | Both derivatives given correctly |
| $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{\sqrt{3}\sec^2\left(t+\frac{\pi}{3}\right)}{\sec t \tan t}$ | M1 A1 | M1: Divides their $\frac{dy}{dt}$ by their $\frac{dx}{dt}$; A1: Correct expression, accept any equivalent |

**(a) Alt:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = \sec t$, $y = \sqrt{3}\frac{\tan t + \sqrt{3}}{1 - \sqrt{3}\tan t}$ | | |
| $\frac{dx}{dt} = \sec t \tan t$ and $\frac{dy}{dt} = \sqrt{3}\frac{(1-\sqrt{3}\tan t)\sec^2 t - (\tan t + \sqrt{3})(-\sqrt{3}\sec^2 t)}{(1-\sqrt{3}\tan t)^2}$ | B1 | |
| $\frac{dy}{dx} = \sqrt{3}\frac{(1-\sqrt{3}\tan t)\sec^2 t - (\tan t+\sqrt{3})(-\sqrt{3}\sec^2 t)}{\sec t \tan t(1-\sqrt{3}\tan t)^2}$ | M1 A1 | |

**Total: (3 marks)**

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## Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $t = \frac{\pi}{3} \Rightarrow x = 2,\ y = -3$ | B1 | Correct values for $x$ and $y$ stated or implied |
| $t = \frac{\pi}{3} \Rightarrow \frac{dy}{dx} = \frac{\sqrt{3}\sec^2\frac{2\pi}{3}}{\sec\frac{\pi}{3}\tan\frac{\pi}{3}} = \ldots (= 2)$ | M1 | Evidence of substituting $t = \frac{\pi}{3}$ into gradient expression; must be evaluated |
| Tangent is $y + 3 = 2(x-2) \Rightarrow y = 2x - 7$ | M1 A1 | M1: Correct method for tangent equation using evaluated gradient; A1cso: Correct equation, must come from correct $\frac{dy}{dx}$ |

**Total: (4 marks)**

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## Part (c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $y = \sqrt{3}\frac{\tan t \pm \tan\frac{\pi}{3}}{1 \pm \tan t \tan\frac{\pi}{3}}$ | M1 | Attempts compound angle formula on $y$; accept sign errors; allow if $\sqrt{3}$ missing; must be seen or used in part (c) |
| $x^2 = \sec^2 t = 1 + \tan^2 t \Rightarrow \tan t = \sqrt{x^2-1} \Rightarrow y = \sqrt{3}\frac{\sqrt{x^2-1}+\sqrt{3}}{1-\sqrt{3}\sqrt{x^2-1}}$ | M1 A1 | M1: Uses $\sec^2 t = 1+\tan^2 t$ to find $\tan t$ in terms of $x$; A1: Any correct Cartesian formula with trig terms evaluated |
| $= \sqrt{3}\frac{\sqrt{x^2-1}+\sqrt{3}}{1-\sqrt{3}\sqrt{x^2-1}} \times \frac{1+\sqrt{3}\sqrt{x^2-1}}{1+\sqrt{3}\sqrt{x^2-1}}$ | M1 | Full method to reach required form: multiplies numerator and denominator by appropriate factor and expands |
| $= \frac{3x^2 + 4\sqrt{3x^2-3}}{4-3x^2}$ | A1 | Achieves correct answer |

**Total: (5 marks)**

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**Alt (c) method 2:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $y\left(1 \pm \sqrt{3}\tan t\right) = \sqrt{3}\tan t \pm 3 \Rightarrow \tan t = \frac{\pm y \pm 3}{\pm y\sqrt{3} \pm \sqrt{3}}$ | M1 A1 | M1: Rearranges to find $\tan t$; uses $\sec^2 t = 1+\tan^2 t$ |
| $x^2 = 1 + \left(\frac{y-3}{\sqrt{3}(y+1)}\right)^2$ | | |
| $\Rightarrow y = \frac{3+\sqrt{3x^2-3}}{1-\sqrt{3x^2-3}} \Rightarrow \frac{3x^2+4\sqrt{3x^2-3}}{4-3x^2}$ | M1 A1 | M1A1 per main scheme |

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**Alt (c) method 3 (sin/cos):**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $y = \sqrt{3}\frac{\sin(t+\frac{\pi}{3})}{\cos(t+\frac{\pi}{3})} = \sqrt{3}\frac{\sin t\cos\frac{\pi}{3}+\cos t\sin\frac{\pi}{3}}{\cos t\cos\frac{\pi}{3}-\sin t\sin\frac{\pi}{3}}$ | M1 | Writes $\tan$ in terms of sin and cos and attempts compound angle formula; accept sign errors |
| $y\left(\cos t - \sqrt{3}\sin t\right) = \sqrt{3}\sin t + 3\cos t \Rightarrow y\left(\frac{1}{x}-\sqrt{3}\sqrt{1-\frac{1}{x^2}}\right) = \sqrt{3}\sqrt{1-\frac{1}{x^2}}+\frac{3}{x}$ | M1 A1 | M1: Applies $\sin^2 t = 1-\cos^2 t$; A1: Any correct Cartesian formula; M1A1 per main scheme for final simplification |