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

Now we substitute the coordinates of the point (0, 1) into the differentiated equation. Substituting gives:

$6(0) - 4(1) \frac{dy}{dx} + 2 - 3\frac{dy}{dx} = 0,$

which simplifies to:

$-4\frac{dy}{dx} - 3\frac{dy}{dx} + 2 = 0.$
Combining the terms yields:

$-7\frac{dy}{dx} + 2 = 0,$
thus,

$\frac{dy}{dx} = \frac{2}{7}.$
This represents the slope of the tangent line.

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus,

$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{\frac{2}{7}} = -\frac{7}{2}.$

Using the point-slope form of a line, we can write the equation of the normal line at the point (0, 1):

$y - y_1 = m(x - x_1),$

substituting in our values:

$y - 1 = -\frac{7}{2}(x - 0)$
which simplifies to:

$y - 1 = -\frac{7}{2} x.$
Rearranging gives us:

$\frac{7}{2} x + y - 1 = 0,$
or multiplying through by 2 to eliminate the fraction:

$7x + 2y - 2 = 0.$
Thus, in the form $ax + by + c = 0$, we can identify a = 7, b = 2, and c = -2.