The shortest distance from the parabola to the straight-line Find the shortest distance from the parabola
$$y^2=64x \tag{1}$$  to the straight-line 
$$4x+3y+46=0\tag{2}$$ I guess, to first find 
$$x=-\frac{3y+46}{4}\tag{3}$$
 and than substitue it into the parabola equation, but this way take to much time, moreover i`m not sure that its right, any hints are welcome
 A: Hint: Two possible ways to resolve: as an extremal problem, finding minimal value; or geometric approach, as a distance between two parallels where one is a tangent of parabola
A: Assume that the distance between the parabola $p$ and the line $\ell$ is the length of a segment $PL$ with $P$ lying on the parabola and $L$ lying on the line. Then, by definition of distance, $PL$ is (one of) the shortest segment(s) joining a point on the parabola with a point on the line. In such a case the Pythagorean theorem ensures $PL\perp\ell$: if it weren't so, it would be possible to move a bit $L$ on $\ell$ and decrease the length of $PL$, contra minimality. For a similar reason (convexity), if $\tau$ is the tangent to the parabola at $P$, we also have $\tau\perp PL$, from which $\tau\parallel\ell$, the unicity of $PL$ and $d(p,\ell)=d(\tau,\ell)=d(P,\ell)$ follow.
In our case the slope of $\ell$ is $-\frac{4}{3}$, hence it is enough to find $P\in p$ such that the slope of $\tau$ is $-\frac{4}{3}$.
Through derivatives we have $P=(9,-24)$ and it is easy to finish:
$$ d(P,\ell) = \frac{\left|4\cdot 9-3\cdot 24+46\right|}{\sqrt{3^2+4^2}}=\color{red}{2}.$$
A: Lines parallel to $(2)$ have the equation
$$4x+3y+d=0\tag{4}$$
where $d$ is an arbitrary real number.
If we intersect $(4)$ and the parabola $(1)$ by elimination $x$ we get the equation
$$y^2-48y+16d=0\tag{5}$$
and further
$$y_{1,2}=-24\pm\sqrt{24^2-16d}\tag{6}$$
A line tangent to a parabola has exactly one intersection point with this parabola. So equation $(5)$ must have only one solution.
From $6)$ we see that happens if 
$$24^2-16d=0\tag{7}$$which means that
$$d=36\tag{8}$$
So the tangent is 
$$ 4x+3y+36=0 \tag{9}$$
Substituting $(8)$ in $(6)$ gives 
$$y=-24 \tag{10}$$
and from $(4)$ we get 
$$x=9 \tag{11}$$
So the tangent $(9)$  touches the parabola at the point $(9,-24)$.
Now find the distance of this point from the line $(2)$ and you are done.
A: Parabola: 
$$y^2=64x\\\frac {dy}{dx}=-\frac {32}y$$
Slope of Line $4x+3y+46=0$ is $-\dfrac 43$. 
A point on Parabola closest to Line is the point at which slope of tangent equals the slope of Line, i.e. $(9,-24)$. 
Hence shortest distance from Parabola to Line is
$$\frac {4(9)+3(-24)+46}{5}=\color{red}2$$
