Equation of Normal I'm struggling to get the same answer as the book on this one. I wonder if someone could please steer me in the right direction?
Q. Find the equation of the normal to $y=x^2 + c$ at the point where $x=\sqrt{c}$
At $x = \sqrt{c}$ then $y = 2c$, so the point given is $(\sqrt{c},2c)$
The gradient of the tangent is $2x$ so the gradient of the normal is $-\frac{1}{2}x$ and the equation of the normal will be $-\frac{1}{2}x + v$
Where $x=\sqrt{c}$ 
then $y = -\frac{1}{2}\sqrt{c} + v$ 
We know $y = 2c$ therefore $2c = -\frac{1}{2}\sqrt{c} + v$ and so 
 $v = 2c + \frac{1}{2}\sqrt{c}$
So the equation of the normal is $y = -\frac{1}{2}x +2c +\frac{1}{2}\sqrt{c}$ 
We can tidy this up to get $2y = -x + 4c + \sqrt{c}$
Unfortunately the book tells me the answer is $2y\sqrt{c} = -x+\sqrt{c}(4c+1)$
It looks like I lost a $\sqrt{c}$ somewhere? Where did I go wrong?
Thank you
Gary
 A: Hint: at the point $x_0=\sqrt{c}$ the slope of the tangent is $2x_0=2\sqrt{c}$, the slope of the normal is $k=-\frac{1}{2x_0}=-\frac{1}{2\sqrt{c}}$. Now use the fact that you know the point coordinates to get the normal equation as $y=kx+m$.
A: You misinterpreted the gradient of tangent: it is  $2x$ at the point with abscissa $x$. At the point with abscissa $\sqrt c$, it is therefor $2\sqrt c$, and the slope of the normal at this same point is $\color{red}{-\frac1{2\sqrt c}}$.
On the other hand, you shouldn't have to solve an equation to find the constant term: you should know the equation of a straight line, given its slope $m$ and the coordinates $(x_0,y_0)$ of a point on the line:
$$y=m(x-x_0)+y_0.$$
In the particular case of a tangent to a curve $y=f(x)$ at a point $(x_0,f(x_0))$, it is
$$y=f'(x_0)(x-x_0)+f(x_0)$$
(i.e. the right-hand side is the Taylor polynomial near $(x_0)$ of degree $1$ near $(x_0)$).
Therefore, an equation of the normal at the point with abscissa $x_0$ is
$$y=-\frac1{f'(x_0)}(x-x_0)+f(x_0).$$
A: Your statement about the gradient of the normal should have been as follows:
The gradient of the tangent is $2x$ so the gradient of the normal is $ \frac {−1}{2x}$ which will be  $ \frac {−1}{2\sqrt c}$
You will get the book's answer with this new slope. 
A: HINT:
The gradient is $2x$ but you have that $x=\sqrt{c}$ so the gradient is equal to $2\sqrt c$
A: Equation of normal is
$$y-y_0=-\frac{1}{f'(x_0)}(x-x_0)$$
In your case as you said $x_0=\sqrt c$ and $y_0=2c$.
$$f'(x_0)=2x_0=2\sqrt c$$
When we put all this back to the equation of normal we get:
$$y-2c=-\frac{1}{2\sqrt c}(x-\sqrt c)$$
$$y=-\frac{1}{2\sqrt c}(x-\sqrt c)+2c$$
$$2y\sqrt c=-x+\sqrt c+4c\sqrt c$$
$$2y\sqrt c=-x+\sqrt c(4c+1)$$
