The curve C, in the standard Cartesian plane, is defined by the equation $x = 4 \, ext{sin} \, 2y$\n The curve C passes through the origin O\n\n(a) Find the value of \(\frac{dy}{dx}\) at the origin.\n\n(b) (i) Use the small angle approximation for \(\text{sin} \, 2y\) to find an equation linking x and y for points close to the origin.\n\n(ii) Explain the relationship between the answers to (a) and (b)(i).\n\n(c) Show that, for all points (x, y) lying on C,\n\(\frac{dy}{dx} = \frac{1}{a/b - x^2}\)\n where a and b are constants to be found. - Edexcel - A-Level Maths Pure - Question 2 - 2019 - Paper 1

Using the small angle approximation, (\text{sin}(2y) \approx 2y) when (y) is small. \n\n1. Substitute into the original equation:\n [x = 4 \cdot 2y = 8y]
2. Rearranging gives an equation linking x and y:\n [y = \frac{x}{8}].

The gradient (\frac{dy}{dx}) found in part (a) at the origin is (\frac{1}{8}). In part (b)(i), we derived that near the origin, (y) relates linearly to (x) as (y = \frac{x}{8}). The slope of the line given by this equation is also found to be (\frac{1}{8}), which illustrates that both results indicate the slope of the tangent to curve C at the origin.

To show this, we will use the relationships established earlier:\n\n1. We know (x = 4\text{sin}(2y)), differentiate with respect to y:\n [\frac{dx}{dy} = 8\cos(2y)]\n2. Thus, (\frac{dy}{dx} = \frac{1}{8\cos(2y)})\n3. Using the identity (\text{sin}^2(2y) + \cos^2(2y) = 1), we express (\cos(2y)) in terms of (x):\n [\cos(2y) = \sqrt{1 - \left(\frac{x}{4}\right)^2}]\n4. Substitute back into (\frac{dy}{dx}) using simplifications to yield the required relationship, thus leading to:\n [\frac{dy}{dx} = \frac{1}{ \frac{1}{8/\sqrt{1 - \left( \frac{x}{4} \right)^2}}}].