Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 an heta, \, y = 4 \, ext{cos}^2 heta, \, 0 \leq \theta < \frac{\pi}{2}$ The point P lies on C and has coordinates (3, 2) - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 7

To find the x-coordinate of the point Q, we start by determining the slope of the curve C at the point P.

1. Compute the derivatives:

   • From the parametric equations, we have: $\frac{dy}{d\theta} = -8\cos(\theta)\sin(\theta)$
   • Also, $\frac{dx}{d\theta} = 3\sec^2(\theta)$
   • Therefore, the slope of the curve at P is: $m = \frac{dy/d\theta}{dx/d\theta} = \frac{-8\cos(\theta)\sin(\theta)}{3\sec^2(\theta)}$

2. Evaluate at P(3, 2):

   • Using the coordinates for P, we find (\theta).
   • From (y = 4 \cos^2 \theta = 2), we find (\theta = \frac{\pi}{4}).
   • Hence, $m = \frac{-8\cos(\frac{\pi}{4})\sin(\frac{\pi}{4})}{3\sec^2(\frac{\pi}{4})} = \frac{-8\cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}{3 \cdot 2} = -\frac{2}{3}$

3. Equation of the normal line:

   • The slope of the normal line is the negative reciprocal: $m_n = \frac{3}{2}$.
   • Using the point-slope form of the line passing through P(3, 2): $y - 2 = \frac{3}{2}(x - 3)$.

4. Find where the normal line cuts the x-axis:

   • Set (y = 0): $0 - 2 = \frac{3}{2}(x - 3) \Rightarrow 2 = \frac{3}{2}(3 - x)$ $4 = 3(3 - x) \Rightarrow 4 = 9 - 3x \Rightarrow 3x = 5 \Rightarrow x = \frac{5}{3}$.

Thus, the x-coordinate of point Q is (\frac{5}{3}).