The curve C has equation x^2 + 2xy - 3y^2 + 16 = 0. (a) Show that dy/dx = (x + y)/(3y - x). (b) Find the equation of the normal to the curve C at the point (2, 2), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

1. M1: Differentiates x^2 term correctly to 2x.
2. M1: Differentiates 2xy using product rule to 2y + 2x(dy/dx).
3. M1: Differentiates -3y^2 term correctly to -6y(dy/dx).
4. A1: Collects terms and correctly rearranges to make dy/dx the subject, leading to the given answer.
5. B1ft: Substitutes (2, 2) into their dy/dx from part (a) to find the gradient of the tangent. (Follow through for minor algebraic errors)
6. M1: Calculates the gradient of the normal using m_normal = -1/m_tangent.
7. M1: Uses y - y1 = m(x - x1) with their normal gradient and point (2, 2).
8. A1: Correct equation of the normal in any valid form.
9. A1: Final answer in the specified form ax + by + c = 0 with integer coefficients. (e.g., 2x - y - 2 = 0 or -2x + y + 2 = 0)