Where do parametric equation for normal line of parametric equation come from? In my calculus II course, I am currently studying parametric curves. And they are awesome! At some point, the book states the following:
if we set parameters x = f(t), and y = g(t),
the parametric equations for the tangent line at time $t_{0}$ are 
x = f ($t_{0}$) + f ' ($t_{0}$)(t - $t_{0}$)
y = g ($t_{0}$) + g ' ($t_{0}$)(t - $t_{0}$)
I understand that these are just the standard line equations, in which f($t_{0}$) is the x coordinate at that time, and g ($t_{0}$) the y coordinate. The derivatives are the slope of the functions at that point.
The next part is what has been puzzling me for a time now
The NORMAL lines at those points are then
x = f ($t_{0}$) + g ' ($t_{0}$)(t - $t_{0}$)
y = g ($t_{0}$) - f ' ($t_{0}$)(t - $t_{0}$) 
I would have expected the slope to change from g ' ($t_{0}$) to 
$\frac{-1}{g ' (t_{0})}$
Am I missing something obvious? Thank you very much for your time!
QUESTION ANSWERED: Thank you very much to all of you, I understand now.
 A: Note that the slope is given by $$\frac{dy}{dx} = \frac{y'(t)}{x'(t)}$$
This should fit your intuition for slope--it's change in $y$ divided by change in $x$. Therefore, in the first formula, the slope is
$$\frac{g'(t_0)}{f'(t_0)}$$
and in the second one, the slope is
$$-\frac{f'(t_0)}{g'(t_0)}$$
which is consistent with your usual formula for the slope of the perpendicular line. 
A: $\frac {dy}{dx} = \frac {\frac {dy}{dt}}{\frac{dx}{dt}}$
With our if we apply this to our tangent lines.
$\frac {dy}{dx} = \frac {g'(t)}{f'(t)}$
and If we apply it to the equations of the normal lines, we get.
$\frac {dy}{dx} = -\frac {f'(t)}{g'(t)}$
which is indeed the negative reciprocal of the our equations for the tangent lines.
However, if you wanted to, you could say:
$x(t) = y(t_0) + \frac 1{f'(t_0)} (t-t_0)\\
y(t) = y(t_0) - \frac 1{g'(t_0)} (t-t_0)$
A: In vector terms, the tangent line has equation
$\begin{bmatrix}
x(t) \\ y(t)
\end{bmatrix} =
\begin{bmatrix}
 f(t_0) \\  g(t_0)
\end{bmatrix} +
(t-t_0)
\begin{bmatrix}
 f'(t_0) \\  g'(t_0)
\end{bmatrix}
= \overrightarrow{p_0} + (t-t_0)\overrightarrow v$
where $\overrightarrow{p_0}$ is the point of tangency and 
$\overrightarrow v$ is the direction vector of the tangent line.
Note that the vector 
$\overrightarrow{v_{\perp}} =\begin{bmatrix}
 g'(t_0) \\  -f'(t_0)
\end{bmatrix}$
is perpendicular to the vector 
$\overrightarrow v = \begin{bmatrix}
 f'(t_0) \\  g'(t_0)
\end{bmatrix}; $ that is to say, their dot product is equal to $0$.
Hence the equation to the line normal to $\begin{bmatrix}
x(t) \\ y(t)
\end{bmatrix}$
at the point $\overrightarrow{p_0} = \begin{bmatrix}
 f'(t_0) \\  g'(t_0)
\end{bmatrix}$ is
$$\begin{bmatrix}
x_\perp(t) \\ y_\perp(t)
\end{bmatrix} =
\begin{bmatrix}
 f(t_0) \\  g(t_0)
\end{bmatrix} +
(t-t_0)
\begin{bmatrix}
 g'(t_0) \\  -f'(t_0)
\end{bmatrix}
= \overrightarrow{p_0} + (t-t_0)\overrightarrow{v_\perp}$$
