multivariable calculus - Confused about directional derivatives

I have the following question:

Calculate the directional derivative of the function at the point and in direction indicated.

$f(x, y) = \arctan(xy)$ at $(1, 2)$ along the line $y = 2x$ in the direction of increasing $x$.

When I looked at the solution, I was confused about the way they solved it:

I understand that we need a gradient vector at (1,2) and some unit vector to give the direction.

I also understand that since it is in the direction of increasing x, x will be positive.
However, how did they get that it is going to be along the line (1,2).
Did they set x = t and then got parametric equations where x = t, and y = 2t?
If so, are the coefficient before t our vector that gives us the direction?

Also, I'm so confused about the use of parametric equations with directional derivatives. Could someone explain relationship between parametric equations and directional derivatives?

Answer

If you are going along the line $y = 2x$ then you find it's directional vector by looking at $\vec{OP}$ where $O = (0,0)$ and $P$ is a point on the line. Take $P = (1,2)$ then the directional vector is $\vec{OP} = \langle1,2 \rangle$.