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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

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    This substitution will compute the intersection of those two. A point on that parabola is of the form $\left[y^2/64, y\right]$ and a distance of a point has a nice formula, so plugging $\left[y^2/64, y\right]$ and finding its minimum (which is a minimum of a quadratic function, hence its stationary point) will give the result. – 2017-02-03
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    From that substitution,I got the equation $y^2+48y+736=0$ where d<0, where i gone wrong? – 2017-02-03
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    As mentioned in my previous comment - this computes the intersection, so it just tells you they don't intersect. – 2017-02-03
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    @pepa.dvorak Ough i totally misunderstand the task, thx – 2017-02-03

4 Answers 4

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Parabola: $$y^2=64x\\\frac {dy}{dx}=-\frac {32}y$$

Slope of Line $4x+3y+46=0$ is $-\dfrac 43$.

Point on Parabola closest to Line is point at which slope of tangent equals slope of Line, i.e. $(9,-24)$.

Hence shortest distance from Parabola to Line is $$\frac {4(9)+3(-34)+46}{5}=\color{red}2$$

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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

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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: $$ d(P,\ell) = \frac{\left|4\cdot 9-3\cdot 24+46\right|}{\sqrt{3^2+4^2}}=\color{red}{2}.$$

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    I see u find a point and compute distance between line and point, but can you pls add more steps on, how to find out that point from parabola, not theory but the derivatives exactly – 2017-02-03
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    All right, that part can be tackled without derivatives, too. Every point on the parabola is associated with a number that gives the slope of the tangent at that point. You may figure out this association by imposing that a discriminant is zero. – 2017-02-03
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    We also have this Lemma dating back to Archimedes: we have a parabola $p$ with vertex at $V$ and some point $P\in p$ with its tangent $\tau$. If $P_1$ is the projection of $P$ on the symmetry axis of the parabola and $P_2$ is the symmetric of $P_1$ with respect to $V$, then $\tau$ goes through $P_2$. – 2017-02-03
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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.