Shortest distance between a point and a parabola.

Question

Find the shortest distance from the point $(1, 0)$ to the parabola $y^2 = 4x$.

As-

Let $D^2=(x-1)^2 + y^2$. Now reduce this to one variable by putting $y^2 = 4x$ to get $D^2 = (x-1)^2+ 4x$. We are working with $D^2$ to make the calculations easier. Clearly $$\frac{d}{dx}D^2 = 2(x-1)+4 = 0 \implies x = -1$$ At $x = -1$, we have $\frac{d^2}{dx^2}D^2 = 2 < 0$ which shows that $D^2$ is minimum. At this $x = -1$.

But this is not a point on parabola. Am i doing some mistake or we have to follow some other method to get the minima?

Answer 1

hint

$$D^2=(x+1)^2$$ $$=(\frac{y^2}{4}+1)^2$$

$D$ is minimal when $y=0$.

Answer 2

Let $\left(\frac{y^2}{4},y\right)$ be a point on our parabola.

Thus, we need to find a minimal value of the following expression. $$\sqrt{\left(\frac{y^2}{4}-1\right)^2+y^2},$$ which is equal to $1$ for $y=0$.

We'll prove that it's a minimal value.

Indeed, we need to prove that: $$\left(\frac{y^2}{4}-1\right)^2+y^2\geq1$$ or $$\frac{y^2}{2}+\frac{y^4}{16}\geq0,$$ which is obvious.