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

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. 
