Figure 3 shows a sketch of the curve C with parametric equations $x = 4 \, ext{cos} \left( t + \frac{\pi}{6} \right), \quad y = 2 \text{sin} t, \quad 0 < t < 2\pi$ (a) Show that $x + y = 2\sqrt{3} \, \text{cos} \, t$ (b) Show that a cartesian equation of C is $(x + y)^2 + a y^2 = b$ where $a$ and $b$ are integers to be determined. - Edexcel - A-Level Maths Pure - Question 7 - 2014 - Paper 7

To show the relationship, we start from the parametric equations:

1. Substitute $x$:
   $x = 4 \text{cos}\left(t + \frac{\pi}{6}\right)$

2. Substitute for $y$:
   $y = 2 \text{sin} \, t$

3. Now, consider: $x + y = 4 \text{cos}\left(t + \frac{\pi}{6}\right) + 2 \text{sin} \, t$

4. We can expand $\text{cos}\left(t + \frac{\pi}{6}\right)$ using the cosine addition formula: $\text{cos}\left(t + \frac{\pi}{6}\right) = \text{cos} \, t \text{cos}\left(\frac{\pi}{6}\right) - \text{sin} \, t \text{sin}\left(\frac{\pi}{6}\right)$

5. Knowing that $\text{cos}\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ and $\text{sin}\left(\frac{\pi}{6}\right) = \frac{1}{2}$, we can write: $\text{cos}\left(t + \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \text{cos} \, t - \frac{1}{2} \text{sin} \, t$

6. Therefore, substituting back: $x + y = 4 \left(\frac{\sqrt{3}}{2} \text{cos} \, t - \frac{1}{2} \text{sin} \, t\right) + 2 \text{sin} \, t$

7. Simplifying this: $= 2\sqrt{3} \text{cos} \, t - 2\text{sin} \, t + 2\text{sin} \, t$ $= 2\sqrt{3} \text{cos} \, t$

Thus, we have shown that:

$x + y = 2\sqrt{3} \text{cos} \, t$