The problem is as follows: Find the point on the parabola $2x=y^{2}$ closest to $(1,0$).
I was highly surprised because I ended up with the correct answer doing something completely different than the solutions manual suggested, and now I am wondering if my method is correct, or if it was simply due to pure luck.
This is how I solved it:
The point in question is (1,0), which means that the parabolic point is in the first quadrant. This way I can avoid implicit differentiation and do the following:
$y=\sqrt{2x} \implies y '(x)= (\sqrt{2x})^{-1} \implies slope\:of\:normal=-\sqrt{2x}$
My logic: The normal of the parabola that crosses (1,0) must also cross the point on the parabola closest to (1,0). Is this logic correct?
Using the point-slope equation to construct the normal line yields:
$y=-x\sqrt{2x}+\sqrt{2x}$
The point is found where the parabola and the normal intersects:
$\sqrt{2x}=-x\sqrt{2x}+\sqrt{2x} \implies x=0 \implies y=0$.
Is this a legitimate method of solving the problem?
Best Answer
Let $(x,y) = \left(\dfrac{y^2}{2},y\right)$ be the point on the parabola that is closest to $(1,0)$, then this point is where the distance of the two point is minimized. $d = \sqrt{\left(\dfrac{y^2}{2} - 1\right)^2 + y^2} = \sqrt{\dfrac{y^4}{4} + 1} \geq 1$, and is minimized when $y = 0$. Thus the sought point is the origin $(0,0)$.