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 as follows:

1. Differentiate term by term: $\frac{d}{dx}(3x^2) - \frac{d}{dx}(y^2) + \frac{d}{dx}(xy) = 0$

2. This gives: $6x - 2y \frac{dy}{dx} + (x \frac{dy}{dx} + y) = 0$

3. Rearranging terms leads to: $6x + y = 2y \frac{dy}{dx} - x \frac{dy}{dx}$ $6x + y = (2y - x) \frac{dy}{dx}$

4. Thus, $\frac{dy}{dx} = \frac{6x + y}{2y - x}$

5. We know the gradient at points P and Q is ( \frac{3}{5} ). So, setting this equal to our derivative: $\frac{6x + y}{2y - x} = \frac{3}{5}$

6. Cross-multiply: $5(6x + y) = 3(2y - x)$ $30x + 5y = 6y - 3x$ $33x - y = 0$

7. Simplifying gives: $y = 33x$

8. We want to show that $y - 2x = 0$: $33x - 2x = 0\Rightarrow y - 2x = 0.$

9. Therefore, at points P and Q, we confirm $y - 2x = 0$ is satisfied.