<s>[INST] Prove Smaller Distance from Hyperbola to Asymptote There is a canonical hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $$ and the asymptote $$ y= \pm \frac{b}{a}x $$ 
Let us say that the value of hyperbola at $x$ is given as
$$y=\frac{b}{a}\sqrt{x^2 - a^2}$$ and the value of asymptote at the same $x$ is given as
$$ y = \frac{b}{a}x$$ 
we will have the difference of hyperbola and asymptote at the same $x$ as
$$ \frac{b}{a}(x-\sqrt{x^2 - a^2})$$
Let us take another point at $x+\delta$ where the difference of hyperbola and asymptote will be
$$ \frac{b}{a}(x+\delta - \sqrt{x^2 - a^2 + 2\delta x+ \delta^2}) $$
Prove that the difference at $x+\delta$ is lower than at $x$.
My attempt is only able to come until this
$$ \sqrt{x^2-a^2+2\delta x+\delta^2}-\delta<\sqrt{x^2-a^2}$$. Then I don't know what to do. Thanks in advance.
 [/INST] OK, by using comment from Blue, I will try to answer the question myself. Since the difference is $$\frac{b}{a}(x-\sqrt{x^2 - a^2}) $$ then the derivative will be $$\frac{b}{a}(1-\frac{x}{\sqrt{x^2-a^2}}) $$ 
Since I am only interested to the value of $x > a$ where $x$ and $a$ is always positive then $$\frac{x}{\sqrt{x^2-a^2}}>1$$ and we always have negative for
$$\frac{b}{a}(1-\frac{x}{\sqrt{x^2-a^2}}) $$ 
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