The curve C has parametric equations $x = 2 \, ext{cos} \, t, \quad y = \sqrt{3} \, \text{cos} \, 2t, \quad 0 \leq t \leq \pi$ (a) Find an expression for $\frac{dy}{dx}$ in terms of $t$ - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 1

To find the coordinates of point P when $t = \frac{2\pi}{3}$:

$x = 2 \cos\left( \frac{2\pi}{3} \right) = 2 \cdot -\frac{1}{2} = -1$

$y = \sqrt{3} \cos\left( 2 \cdot \frac{2\pi}{3} \right) = \sqrt{3} \cdot \cos\left( \frac{4\pi}{3} \right) = \sqrt{3} \cdot -\frac{1}{2} = -\frac{\sqrt{3}}{2}$

Thus, the coordinates of P are ((-1, -\frac{\sqrt{3}}{2})).

The gradient of line l, being the negative reciprocal of the gradient of C at P, is:

The existing gradient is $2\sqrt{3}$, hence:

$\text{slope of l} = -\frac{1}{2\sqrt{3}}$

Using the point-slope form, we have:

$y - (-\frac{\sqrt{3}}{2}) = -\frac{1}{2\sqrt{3}}(x - (-1))$

After simplification,

$2y + \sqrt{3} = -\frac{1}{2\sqrt{3}}x - \frac{1}{2\sqrt{3}}$

Thus, multiplying through by $2\sqrt{3}$ gives:

$4\sqrt{3}y + 3 = -x - 1$

Rearranging gives:

$2x - 2\sqrt{3}y - 1 = 0$

Substituting $x = 2 \cos t$ and $y = \sqrt{3} \cos 2t$ into $2x - 2\sqrt{3}y - 1 = 0$:

$2(2 \cos t) - 2\sqrt{3} (\sqrt{3} \cos 2t) - 1 = 0$

This simplifies to:

$4 \cos t - 6 \cos 2t - 1 = 0$

Using the identity $\cos 2t = 2\cos^2 t - 1$, we rewrite the equation:

$4 \cos t - 6(2\cos^2 t - 1) - 1 = 0$

which translates to:

$-12\cos^2 t + 10\cos t - 5 = 0$

Factoring or using the quadratic formula:

Let $u = \cos t$. Solving:

$12u^2 - 10u + 5 = 0$

We apply the quadratic formula:

$u = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 12 \cdot 5}}{2 \cdot 12} = \frac{10 \pm \sqrt{100 - 240}}{24}$

After simplification, we find:

$u =\frac{5}{6} \text{ or }\frac{-1}{3}$

Choosing $\cos t = \frac{5}{6}$ (the valid value), we then find:

Consequently:

$x = 2 \cdot \frac{5}{6} = \frac{5}{3}, \quad y = \sqrt{3} \cdot \cos(2t) = \sqrt{3} \cdot \sqrt{1 - \left(\frac{5}{6}\right)^2}$

Thus, substituting to find the coordinates of Q gives:

$Q = \left(\frac{5}{3}, \frac{7}{18 \sqrt{3}}\right)$