I'm reading Binmore and Davies, Calculus Concepts and Methods. On page 105, there's a function $$\tau(x,y) = 32 - 3\ln(1 + x^2 + 4y^2).$$ The gradient is calculated as $$\left(\frac{-3 \cdot 2x}{1 + x^2 + 4y^2}, \frac{-3\cdot 8y}{1 + x^2 + 4y^2} \right)^T$$
The text then states that the gradient vector has direction $(-x, -4y)^T$.
Where did this $(-x, -4y)$ come from? The text evaluates the gradient at the point $(x, y)^T = (10, 10)^T$ which results in $(\frac{-60}{501}, \frac{-240}{501})$. The text refers to the direction of this as $(-1, -4)^T$.
I thought that the direction might just be the unit vector of the gradient but the length of $(-1, -4)^T$ is not equal to 1.
The other questions on here related to the direction of the gradient have to do with maximum/minimum steepness or the direction being reported as a unit vector. Appreciate any help.
Note that $$\nabla \tau(x,y)=\left(\frac{-3 \cdot 2x}{1 + x^2 + 4y^2}, \frac{-3\cdot 8y}{1 + x^2 + 4y^2} \right)^T=\frac{6}{1 + x^2 + 4y^2}(-x, -4y)^T$$ Since the scalar $\frac{6}{1 + x^2 + 4y^2}$ is a positive real number, then the gradient $\nabla \tau(x,y)$ and the vector $(-x, -4y)^T$ are pointing to the same direction.
The same can be said for $$\left(\frac{-60}{501}, \frac{-240}{501}\right)=\frac{60}{501}(-1, -4)^T.$$