Q:

Minimum distance problem

Find the coordinates of the point in the graph of $f(x)=2x^2+3$ that is closer to the point $(5,-1)$

To start off, I found the first order derivative of the function so I could get the slope of the tangent line to $f(x)$
$$f´(x)=4x$$
Then, I needed to find the equation of the perpendicular line that goes through $(5,-1)$.
If $m_t*m_n=-1,$ therefore $m_n=-{1 \over4x}$
And then I proceeded to find the equation of the perpendicular line, getting as a result:
$$y=-\frac 54+ {5 \over 4x} $$
But as you can see, that's not a line. So I'm asking, what am I doing wrong? 
I'm thinking I have it wrong since I stated the slope of the tangent line is $4x$, I guessing I need a number.

A:

You need to find a distance between any given point on parabola and $(5, 1)$ and find it's minimum with respect to $x$. So 
$$d^2(x) = (x - 5)^2 + (2x^2 + 3 + 1)^2 = x^2-10x + 25 + 4(x^4 + 4x^2 + 4) = 4x^4 + 17x^2-10x+41$$
$$\left[
d^2(x)
\right]' = 16x^3 + 34x - 10 = 0$$
Latter cubic equation has roots, but it looks like they're not rational, or I messed up somewhere with numbers. But idea holds anyway.
But if you want to stick to your method, you need to:
$
\displaystyle a)\ k = -\frac 1{4x_0}\\
\displaystyle b)\ -1 = -\frac 5{4x_0} + b \quad \Rightarrow \quad b = -1 + \frac 5{4x_0} \\
\displaystyle c)\ 2x_0^2+3=-\frac 14 - 1 + \frac 5{4x_0} \\
\displaystyle d)\ 8x_0^2 + 17x_0-5=0
$
which is essentially the same equation.

