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

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

$x = 2 \sin y \implies \frac{dx}{dy} = 2 \cos y.$

Using the chain rule, we have:

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

Next, we evaluate this expression at the point $P \left(\sqrt{2}, \frac{\pi}{4}\right)$:

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

Therefore:

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

The gradient of the normal line at point $P \left(\sqrt{2}, \frac{\pi}{4}\right)$ can be found by taking the negative reciprocal of (\frac{dy}{dx}):

$m = -\frac{1}{\frac{1}{\sqrt{2}}} = -\sqrt{2}.$

Using the point-slope form of a line:

$y - y_1 = m(x - x_1),$

where $y_1 = \frac{\pi}{4}$ and $x_1 = \sqrt{2}$, we substitute to get:

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

Now we rearrange this to the form $y = mx + c$:

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

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

Thus, the equation of the normal line is:

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