If you have a hyperbola defined by the equation : $(x^2/a^2) - (y^2/b^2) = 1$. What would be the range of possible values for the gradients of all the tangents with this hyperbola which are parallel to the line $y=kx$.
Logically, $|k| > b/a$ given that $y=\pm b/a$ are the asymptotes of the hyperbole.
So the gradient can't be smaller than $b/a$ because then the line of the "tangent" would intersect the hyperbola in more than one point and it can't be equal to $b/a$ because it wouldn't intersect the hyperbola at all.
My question is, how would you prove this in a mathematical way. Should you use the rule of the tangent to the hyperbola? $(a^2*k^2 - b^2 = t^2$, where the tangent is given by: $y=kx + t$) I tried using that but I couldn't really get a valid solution.
You’re almost there. You’ve got an equality that relates the slope $m$ of a tangent line to its $y$-intercept $t$: $$t^2=m^2a^2-b^2.$$ (I’ve changed the $k$ in your question to the more usual $m$.) This immediately gives you some bounds on $m$: since $t^2\ge0$, we must also have $m^2\le{b^2\over a^2}$. As you’ve already observed, equality produces an asymptote, so the inequality must be strict.