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 the derivative of $y$ with respect to $x$, we differentiate both sides of the equation $x = 2 \, \text{sin} \, y$:

$\frac{dx}{dy} = 2 \cos y \Rightarrow \frac{dy}{dx} = \frac{1}{2 \cos y}.$

Next, we evaluate this at the point $P$ where $y = \frac{\pi}{4}$:

$\cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}},$

thus:

$\frac{dy}{dx} = \frac{1}{2 \cdot \frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}.$

This shows that $\frac{dy}{dx} = \frac{1}{\sqrt{2}}$ at P.

The slope of the tangent line at point $P$ is given by:

$m_t = \frac{dy}{dx} = \frac{1}{\sqrt{2}}.$

The slope of the normal line is the negative reciprocal:

$m_n = -\frac{1}{m_t} = -\sqrt{2}.$

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

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

To express this in the form $y = mx + c$, we rewrite:

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

which simplifies to:

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

Thus, the equation of the normal line is:

$y = -\sqrt{2}x + \left(2 + \frac{\pi}{4}\right).$