A curve C is described by the equation $$3x^4 + 4y^2 - 2x + 6y - 5 = 0.$$ Find an equation of the tangent to C at the point (1, -2), giving your answer in the form $ax + by + e = 0$, where a, b and c are integers. - Edexcel - A-Level Maths Pure - Question 3 - 2006 - Paper 7

Next, substitute $x = 1$ and $y = -2$ into the derived expression:

$\frac{dy}{dx} = \frac{2 - 6(1) - 6(-2)}{6(1) + 8(-2)} = \frac{2 - 6 + 12}{6 - 16} = \frac{8}{-10} = -\frac{4}{5}.$

Now, we can use the point-slope form to find the tangent line:

$y + 2 = -\frac{4}{5}(x - 1).$

This simplifies to:

$y + 2 = -\frac{4}{5}x + \frac{4}{5}$

or

$y = -\frac{4}{5}x + \frac{4}{5} - 2 = -\frac{4}{5}x - \frac{6}{5}.$

To express this in the required form $ax + by + e = 0$, we rearrange:

$\frac{4}{5}x + y + \frac{6}{5} = 0$

Multiplying through by 5 to eliminate the fraction gives:

$4x + 5y + 6 = 0.$