The curve C has equation $$x^2 - 3xy - 4y^2 + 64 = 0$$ (a) Find \(\frac{dy}{dx}\) in terms of x and y - Edexcel - A-Level Maths Pure - Question 4 - 2015 - Paper 4

For (\frac{dy}{dx} = 0), we set the numerator equal to zero:

$2x - 3y = 0$

From this, we can express y in terms of x:

$y = \frac{2}{3}x$

Next, we substitute this into the original equation of the curve:

$x^2 - 3x\left(\frac{2}{3}x\right) - 4\left(\frac{2}{3}x\right)^2 + 64 = 0$

Simplifying:

$x^2 - 2x^2 - \frac{16}{9}x^2 + 64 = 0$

Combining like terms:

$\left( 1 - 2 - \frac{16}{9} \right)x^2 + 64 = 0$

Converting all terms to a common denominator:

$\left(-\frac{9}{9} - \frac{18}{9} - \frac{16}{9}\right)x^2 + 64 = 0$

This reduces to:

$-\frac{43}{9}x^2 + 64 = 0$

Solving for x:

$x^2 = \frac{64 \cdot 9}{43} = \frac{576}{43} \Rightarrow x = \pm \sqrt{\frac{576}{43}}$

Now substitute these x-values back to find the corresponding y-values:

$y = \frac{2}{3}\sqrt{\frac{576}{43}}$

Thus, the coordinates of the points are:

$\left( \sqrt{\frac{576}{43}}, \frac{2}{3}\sqrt{\frac{576}{43}} \right) \text{ and } \left( -\sqrt{\frac{576}{43}}, -\frac{2}{3}\sqrt{\frac{576}{43}} \right).$