Equation of Normal? I got a question that I need help on:
Q1: The curve C has the equation $2x+3y^2+3x^2y=4x^2$.
The point P on the curve has coordinates $(-1,1).$
(a): Find the gradient of the curve at P.
So I did: $\frac{d}{dx}(2x+3y^2+3x^2y=4x^2$
$2+6y\frac{dy}{dx}+6xy+3x^2\frac{dy}{dx}=8x$
$\frac{dy}{dx}=\frac{8x-2}{6y+6xy}$
Using the coordinates (-1,1)
$\frac{dy}{dx}=\frac{1}{2}$
Q2: Hence find the equation of the normal to curve C at P, giving your answer in the form of $ax+by+c=0$.
I just don't understand what they want me to do, "normal to the curve C at P?"
 A: $2x+3y^2+3x^2y = 4x^2$
$2+6yy'+6xy+3x^2y'=8x$
$y' = \frac{8x-2-6xy}{6y+3x^2}$
$y'|_P = \frac{-8-2+6}{6+3} =\frac{-4}{9} $

Slope of the normal at $P$,  $\ \ $ $m = -\frac{1}{y'} = +\frac{9}{4}$

So the equation of the normal to the curve at $P\equiv(-1,1)$ is,
$y - 1 = m(x-(-1))$
$y - 1 = \frac{9}{4}(x+1)$
$4y-4 = 9x + 9$

$$9x-4y+13=0$$

A: 
$2+6y\frac{dy}{dx}+6xy+3x^2\frac{dy}{dx}=8x$
$\frac{dy}{dx}=\frac{8x-2}{6y+6xy}$

That didn't go right, solving for $\frac{dy}{dx}$ gives:
$$\frac{dy}{dx}=\frac{8x-2-6xy}{6y+3x^2}$$
which evaluates to $-\tfrac{4}{9}$ at $(-1,1)$.


I just don't understand what they want me to do, "normal to the curve C at P?"

Now a normal (line) is perpendicular to the curve and thus to the tangent line, which means you want a slope corresponding to a line which is orthogonal to one with slope $-\tfrac{4}{9}$, that is $\tfrac{9}{4}$.
The problem is now reduced to setting up the equation of a line through the point $(-1,1)$ and with slope $\tfrac{9}{4}$:
$$y-1=\tfrac{9}{4}\left(x+1\right)$$
