Shortest Distance to Point from Parabola.

Question

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.

Answer 1

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 $x=0$.

Answer 2

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 $1$.

Answer 3

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 $PQ=1$.

Answer 4

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.