A curve C is described by the equation $$3x^4 + 4y^3 - 2x + 6y - 5 = 0.$$ 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. - Edexcel - A-Level Maths Pure - Question 4 - 2006 - Paper 7

Now, we need to evaluate the derivative at the point (1, -2).

Substituting $x = 1$ and $y = -2$ into the expression: $\frac{dy}{dx} = \frac{2 - 6(1)}{12(-2)^2 + 6} = \frac{2 - 6}{48 + 6} = \frac{-4}{54} = -\frac{2}{27}.$
Thus, the gradient at the point (1, -2) is $-\frac{2}{27}$.

We use the point-slope form of the line equation: $y - y_1 = m(x - x_1),$
where $(x_1, y_1) = (1, -2)$ and $m = -\frac{2}{27}$.
Substituting these values gives: $y + 2 = -\frac{2}{27}(x - 1).$
Expanding this: $y + 2 = -\frac{2}{27}x + \frac{2}{27}.$
Rearranging gives: $\frac{2}{27}x + y + 2 - \frac{2}{27} = 0.$
Multiplying through by 27 to eliminate the fraction: $2x + 27y + 54 - 2 = 0$
yielding: $2x + 27y + 52 = 0.$