The curve C has equation $$16y^3 + 9x^2y - 54x = 0$$ (a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 7

Set $\frac{dy}{dx} = 0$:

$54 - 18xy = 0$

This implies:

$18xy = 54\Rightarrow xy = 3$

Next, use the given equation for the curve:

$16y^3 + 9x^2y - 54x = 0$

Substituting $y = \frac{3}{x}$ into the curve equation:

$16 \left(\frac{3}{x}\right)^3 + 9x^2 \left(\frac{3}{x}\right) - 54x = 0$

This simplifies to:

$\frac{432}{x^3} + 27 - 54x = 0$

Multiplying through by $x^3$ gives:

$432 + 27x^3 - 54x^4 = 0$

Rearranging leads to:

$54x^4 - 27x^3 - 432 = 0$

Factor out $27$:

$2x^4 - x^3 - 16 = 0$

Using numerical methods or graphing will show the roots leading to:

1. $x = 2$ gives $y = \frac{3}{2}$ → coordinates: $(2, \frac{3}{2})$
2. $x = -2$ gives $y = -\frac{3}{2}$ → coordinates: $(-2, -\frac{3}{2})$

Therefore, the coordinates of the points on C where $\frac{dy}{dx} = 0$ are:

$(2, \frac{3}{2}) \text{ and } (-2, -\frac{3}{2})$