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 3 - 2015 - Paper 4

Setting the numerator of $\frac{dy}{dx}$ equal to zero gives:

$3y - 2x = 0$

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

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

We will substitute this into the original equation to find the coordinates:

Substituting $y = \frac{2}{3}x$ into:

$x^2 - 3xy - 4y^2 + 64 = 0$

We have:

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

Simplifying this:

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

This gives:

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

Combining the x terms:

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

$\left(-\frac{25}{9}\right)x^2 + 64 = 0$

Solving for x gives:

$x^2 = \frac{64 \cdot 9}{25} = \frac{576}{25}\implies x = \pm \frac{24}{5}$

Now substituting back to find y:

When $x = \frac{24}{5}$:

$y = \frac{2}{3}\left(\frac{24}{5}\right) = \frac{48}{15} = \frac{16}{5}$

When $x = -\frac{24}{5}$:

$y = \frac{2}{3}\left(-\frac{24}{5}\right) = -\frac{48}{15} = -\frac{16}{5}$

Thus, the points are:

$\left(\frac{24}{5}, \frac{16}{5}\right)\quad \text{and} \quad \left(-\frac{24}{5}, -\frac{16}{5}\right)$