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Given the curve $y^2 = 4x$, find the point on the parabola that is the shortest distance to the point $(1,0)$

My work:

\begin{align}d &= \sqrt{(x-1)^2 + y^2}\\ d &= \sqrt{(x-1)^2 +4x}\\ d &= \sqrt{x^2 - 2x + 1 + 4x}\\ d &= \sqrt{x^2 + 2x +1}\\ d &= x + 1\end{align}

At this point, taking the derivative would be useless. So, my question is: Is there a way using calculus that I can solve the above problem.

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    $\begingroup$ Hint for general problems like these: the point on the curve that minimizes the function giving the distance between the curve and your given point will be the same point that minimizes the square of this distance function. In other words, the point that minimizes $d$ is precisely the point that minimizes $d^2$. And this is convenient because you can avoid getting bogged down with the square root. For instance, see here for a worked example. $\endgroup$ – Kaj Hansen Feb 22 '19 at 14:02
  • $\begingroup$ Of course, the square root didn't cause much hassle in this problem. But it could've, given a problem with a different curve. $\endgroup$ – Kaj Hansen Feb 22 '19 at 14:04
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Continuing your computation, we have

$$d=|x+1|.$$

But from $y^2=4x$ we draw that $x\ge0$ so that

$$|x+1|\ge1$$ and the equality is achieved by $x=0$.

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Why do you want to use Calculus? It follows from your computations that the shortest distance is reached when $x=0$ and that that distance is $1$.

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  • $\begingroup$ I'm sorry, but how does it follow that $x = 0$? Is there a proper way to solve for $x$? $\endgroup$ – S. Sharma Feb 22 '19 at 13:48
  • $\begingroup$ Since $4x=y^2\geqslant0$, you know that $x\geqslant0$. And, since $x\in[0,\infty)$ and since $d=x+1$, the minimal value that $d$ can take is $1$, which is obtained when $x=0$. $\endgroup$ – José Carlos Santos Feb 22 '19 at 13:50
  • $\begingroup$ That makes sense, thank you. $\endgroup$ – S. Sharma Feb 22 '19 at 13:52
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Let $P(p^2, 2p)$ be a point on the parabola, and let $Q = (1,0)$. Then $PQ = \sqrt{(p^2 - 1)^2 + (2p-0)^2} = \sqrt{p^4 - 2p^2 + 1 + 4p^2} = \sqrt{(p^2 +1)^2} = p^2+1$.

Thus, $PQ = p^2 + 1$, so the minimum value of $PQ$ occurs at $p=0$. Thus, $P=(0,0)$, with $PQ=1$.

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  • $\begingroup$ It seems we performed similar steps. Why did my steps not reach a solution but yours did? $\endgroup$ – S. Sharma Feb 22 '19 at 13:45
  • $\begingroup$ @S.Sharma Your steps were close to the solution, but you did not check the range of $x$ (which in this question is $x \geq 0$, as $y^2 \geq 0$). $\endgroup$ – Sharky Kesa Feb 22 '19 at 13:47
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You are right: $d=x+1$, but $x\ge0$. Hence the minimum is $d=1$ for $x=0$.

By the way: $(1,0)$ is the focus of the parabola, hence $d=x+1$ follows from the focus-directrix property.

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