C4 January 2006 Q1
1. A curve \(C\) is described by the equation
\[3x^2 + 4y^2 - 2x + 6xy - 5 = 0.\]
Find an equation of the tangent to \(C\) at the point \((1, -2)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. (7)
| Scheme | Marks |
|---|---|
| Differentiates to obtain: \(6x + 8y\frac{\mathrm{d}y}{\mathrm{d}x} - 2,\) | M1 A1, |
| \(\ldots\ldots\ldots\ldots + \left(6x\frac{\mathrm{d}y}{\mathrm{d}x} + 6y\right) = 0\) | +(B1) |
| \(\left[\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2 - 6x - 6y}{6x + 8y}\right]\) | |
| Substitutes \(x = 1, y = -2\) into expression involving \(\frac{\mathrm{d}y}{\mathrm{d}x}\), to give \(\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{8}{10}\) | M1, A1 |
| Uses line equation with numerical ‘gradient’ \(y - (-2) = (\text{their gradient})(x - 1)\) or finds \(c\) and uses \(y = (\text{their gradient})\,x + \text{“}c\text{”}\) | M1 |
| To give \(5y + 4x + 6 = 0\) (or equivalent \(= 0\)) | A1ft |
| (7) | |
| (7 marks) |