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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.

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  • $\begingroup$ Your approach is a reasonable one. Keep at it. You have an equation that gives a relation between the slope and $y$-intercept of the possible tangent lines. Can you derive an inequality involving only $k$ from it? $\endgroup$ – amd Feb 4 '18 at 20:29
  • $\begingroup$ I am probably to stupid or too tired to be able to find anything. I tried various different things, but I can't come up with anything because I don't have a "condition". A rule of some sort to which the gradient has to obey. Like for example "K > 0". Unless I find something like that, I cannot prove this, not with my current knowledge. (I am a High-School student so I have much to learn) $\endgroup$ – Teabx Feb 4 '18 at 20:40
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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.

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  • $\begingroup$ Oh my god, how did I miss that. Thank you so much! $\endgroup$ – Teabx Feb 5 '18 at 5:49

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