A curve C has parametric equations $x = ext{sin}^2 t, \quad y = 2 \text{tan} t, \quad 0 < t < \frac{\pi}{2}$ (a) Find $\frac{dy}{dx}$ in terms of $t$ - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 6

1. At $t = \frac{\pi}{3}$, we calculate: $x = \sin^2\left(\frac{\pi}{3}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$ $y = 2 \tan\left(\frac{\pi}{3}\right) = 2 \cdot \sqrt{3}$

2. The slope of the tangent line is given by $\frac{dy}{dx}$ at $t = \frac{\pi}{3}$: $\frac{dy}{dx} = \frac{\sec^2\left(\frac{\pi}{3}\right)}{\sin\left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}\right)}$ Calculating each term: $\sec^2\left(\frac{\pi}{3}\right) = 4, \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ So, $\frac{dy}{dx} = \frac{4}{\frac{\sqrt{3}}{2} \cdot \frac{1}{2}} = \frac{16}{\sqrt{3}}$

3. The equation of the tangent line at point $P$ is $y - 2\sqrt{3} = \frac{16}{\sqrt{3}}(x - \frac{3}{4})$. Set $y = 0$ to find the x-intercept: $0 - 2\sqrt{3} = \frac{16}{\sqrt{3}}(x - \frac{3}{4})$ Simplifying, $x - \frac{3}{4} = -\frac{\sqrt{3}}{8}$ Thus, $x = \frac{3}{4} - \frac{\sqrt{3}}{8} = \frac{3}{4} - \frac{\sqrt{3}}{8}$ Finding a common denominator gives: $x = \frac{6}{8} - \frac{\sqrt{3}}{8} = \frac{6 - \sqrt{3}}{8}$

Thus, the x-coordinate of point P is: $x_P = \frac{3}{8}$