The curve C has equation $x = 2 \, ext{sin} \, y.$ (a) Show that the point $P \left( \sqrt{2}, \frac{\pi}{4} \right)$ lies on C - Edexcel - A-Level Maths Pure - Question 3 - 2007 - Paper 6

To find $\frac{dy}{dx}$, we first differentiate the equation of the curve with respect to $y$:

1. Starting from the equation $x = 2 \, \text{sin} \, y$,

2. Differentiate both sides: $\frac{dx}{dy} = 2 \, \text{cos} \, y.$

3. Hence, we find: $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{2 \, \text{cos} \, y}.$

4. At the point $y = \frac{\pi}{4}$, we have: $\text{cos} \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}.$

5. Substituting into the derivative gives: $\frac{dy}{dx} = \frac{1}{2 \cdot \frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2}}.$

First, we determine the slope of the normal line at point P, which is the negative reciprocal of the derivative found in part (b):

1. The slope of the tangent line at P is $\frac{1}{\sqrt{2}}$, thus the slope of the normal line is: $m = -\sqrt{2}.$

2. Next, we use the point-slope form of the line, where the equation of the normal line is given by: $y - y_1 = m (x - x_1)$ where $P = \left( \sqrt{2}, \frac{\pi}{4} \right)$, giving: $y - \frac{\pi}{4} = -\sqrt{2} \left( x - \sqrt{2} \right).$

3. Rearranging to find the standard form: $y = -\sqrt{2}x + 2 + \frac{\pi}{4}.$ This can be expressed in the form $y = mx + c$, where $m = -\sqrt{2}$ and $c = 2 + \frac{\pi}{4}$.