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 5 - 2007 - Paper 6

We start by finding rac{dy}{dx} using implicit differentiation on the equation $x = 2 \sin y$.

Differentiating both sides with respect to $x$, we get:

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

Rearranging this gives:

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

Now, we substitute $y = \frac{\pi}{4}$, for which:

$\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}.$

Thus, substituting into our expression yields:

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

To find the equation of the normal line at point $P \left( \sqrt{2}, \frac{\pi}{4} \right)$, we first determine the slope of the normal. The slope of the tangent at P is given by $m_t = \frac{dy}{dx} = \frac{1}{\sqrt{2}}$, hence the slope of the normal line, $m_n$, is the negative reciprocal:

$m_n = -\sqrt{2}.$

Using the point-slope form of the line's equation:

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

where $(x_1, y_1) = \left( \sqrt{2}, \frac{\pi}{4} \right)$, we have:

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

Expanding this, we arrive at:

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

We can write this in the standard form $y = mx + c$ where $m = -\sqrt{2}$ and $c = 2 + \frac{\pi}{4}$.