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 5 - 2007 - Paper 6

To find ( \frac{dy}{dx} ), we can differentiate the curve's equation implicitly:

1. Start with the equation: $x = 2 \sin y$.
2. Differentiating both sides with respect to $x$: $1 = 2 \cos y \frac{dy}{dx}.$
3. Solving for ( \frac{dy}{dx} ): $\frac{dy}{dx} = \frac{1}{2 \cos y}.$
4. At the point $P$ where $y = \frac{\pi}{4}$: $\frac{dy}{dx} = \frac{1}{2 \cos \left( \frac{\pi}{4} \right)} = \frac{1}{2 \cdot \frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2}}.$

Thus, the result is verified.

The slope of the normal line is the negative reciprocal of the tangent slope:

$m_{normal} = - \frac{1}{\frac{dy}{dx}} = - \sqrt{2}.$

Using the point-slope form of the line equation, $y - y_1 = m (x - x_1)$, with point $P \left( \sqrt{2}, \frac{\pi}{4} \right)$:

1. Substitute the point and slope: $y - \frac{\pi}{4} = -\sqrt{2} \left( x - \sqrt{2} \right).$
2. Rearranging gives: $y = -\sqrt{2} x + 2 + \frac{\pi}{4}.$

Thus the equation of the normal line is:

$y = -\sqrt{2} x + 2 + \frac{\pi}{4}.$