Let us say that a line is defined as $$ y=\frac{\sqrt{c^2-a^2}}{a}x $$ and $a<c$. Can we tell when the value of $a$ becomes insignificant with respect to $c$?
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It is not clear what you mean by "insignificant." However, observe that $ \dfrac{\sqrt{c^2 - a^2}}{a}= \sqrt{\dfrac{c^2 - a^2}{a^2}}= \sqrt{\left(\dfrac{c}{a}\right)^2 - 1} $, which is probably a step towards whatever it is you mean. – mathematics2x2life Feb 20 '21 at 08:45
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I mean the value of $a$ that small enough to consider it as infinity. – Semar Feb 20 '21 at 08:50
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Again, that makes no sense. I assume you mean what $a$ makes this as large as you'd like. Certainly $\lim_{a \to 0^+} \sqrt{(c/a)^2-1}= \infty$. If you wanted to make this explicit, simply fix a value $c$ and solve the inequality $\sqrt{(c/a)^2-1} > M$ for $a$, where $M$ is the size that you want the term to be. – mathematics2x2life Feb 20 '21 at 09:00
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You probably mean that if we keep increasing $c$ and keep $a$ constant, when can we ignore $a$. This is usually done in science where we need to approximate things and it us upto us what degree of accuracy we want. There is no such standard where we can say that $c >>a$.
However, suppose if $c=100$ and $a=1$, for calculation purposes, we may simply write your expression as $$y=\dfrac{\sqrt{10000-1}}{1}x\approx \sqrt{10000}x=100x$$
Which differs from actual answer by roughly $0.005x.$ Now it is upto you to accept it or not.
V.G
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