A curve has equation $$x^2 + 2xy - 3y^3 + 16 = 0.$$ Find the coordinates of the points on the curve where \(\frac{dy}{dx} = 0\. - Edexcel - A-Level Maths Pure - Question 4 - 2005 - Paper 6

Rearranging the equation to isolate (\frac{dy}{dx}):

$$(2x + 2y) + (2x - 9y^2)\frac{dy}{dx} = 0 \Rightarrow (2x - 9y^2)\frac{dy}{dx} = - (2x + 2y) \Rightarrow \frac{dy}{dx} = \frac{-(2x + 2y)}{(2x - 9y^2)}.$$

To find the coordinates where (\frac{dy}{dx} = 0), set the numerator to zero:

$$-(2x + 2y) = 0 \Rightarrow 2x + 2y = 0 \Rightarrow y = -x.$$

Now substitute (y = -x) back into the original curve equation:

$$x^2 + 2x(-x) - 3(-x)^3 + 16 = 0 \Rightarrow x^2 - 2x^2 + 3x^3 + 16 = 0 \Rightarrow 3x^3 - x^2 + 16 = 0.$$

Solving this cubic equation for (x) yields:

1. Evaluating specific values, we find two solutions:
   • When (x = 2), (y = -2)
   • When (x = -2), (y = 2)

Thus, the coordinates of the points on the curve where (\frac{dy}{dx} = 0) are ((2, -2)) and ((-2, 2)).