A curve has equation $3x^2 - y^2 + xy = 4$ - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 7

To differentiate the equation $3x^2 - y^2 + xy = 4$ implicitly with respect to $x$, we apply the product rule and chain rule:

1. Differentiate each term:

   • The derivative of $3x^2$ is $6x$.
   • The derivative of $-y^2$ is $-2y \frac{dy}{dx}$ (using chain rule).
   • The derivative of $xy$ is $x \frac{dy}{dx} + y$ (using product rule).

   Putting it all together, we get:

   $6x - 2y \frac{dy}{dx} + \left( x \frac{dy}{dx} + y \right) = 0$

2. Rearranging gives:

   $6x + y - 2y \frac{dy}{dx} + x \frac{dy}{dx} = 0$

   Combine terms with $\frac{dy}{dx}$:

   $6x + y = \left(2y - x\right) \frac{dy}{dx}$

3. Therefore,

   $\frac{dy}{dx} = \frac{6x + y}{2y - x}$

4. To find the points P and Q where the gradient rac{dy}{dx} equals rac{3}{5}, set:

   $\frac{6x + y}{2y - x} = \frac{3}{5}$

5. Cross-multiplying gives:

   $5(6x + y) = 3(2y - x)$

   Expanding:

   $30x + 5y = 6y - 3x$

   Combine like terms:

   $33x - y = 0$ or $y = 33x$

6. Substituting $y = 33x$ into the original equation:

   $3x^2 - (33x)^2 + x(33x) = 4$

   This simplifies to:

   $3x^2 - 1089x^2 + 33x^2 = 4$

   $-1053x^2 = 4$

7. Solving for $x$, we find:

   $x^2 = \frac{-4}{-1053} \Rightarrow x = \pm \sqrt{\frac{4}{1053}}$ Therefore, $y = 33 \left(\pm \sqrt{\frac{4}{1053}}\right)$

   At points P and Q, we have $y - 2x = 0$, therefore confirming the relation.