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The angle between the two asymptotes of the hyperbola $\dfrac{{{x}^{2}}}{25}-\dfrac{{{y}^{2}}}{16}=1$ is
A. ${{\tan }^{-1}}\dfrac{4}{5}$
B. 2${{\tan }^{-1}}\dfrac{4}{5}$
C. 2${{\tan }^{-1}}\dfrac{5}{4}$
D. $\pi -2{{\tan }^{-1}}\dfrac{4}{5}$

Last updated date: 17th Jul 2024
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Hint: See the question and compare the equation with the general equation of a hyperbola. You will get the value of$a$and$b$. Use$y=\dfrac{b}{a}x$and$y=-\dfrac{b}{a}x$ then use it$\tan \theta =\dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}$. You will get the answer. Try it.

A hyperbola is a type of smooth curve lying in a plane, defined by its geometric properties or by equations for which it is the solution set.

A hyperbola has two pieces, called connected components or branches, that are mirror images of each other and resemble two infinite bows.
The hyperbola is one of the three kinds of conic section, formed by the intersection of a plane and a double cone.

An asymptote to a curve is a straight line, to which the tangent to the curve tends as the point of contact goes to infinity.

Asymptotes are imaginary lines that a function will get very close to, but never touch. The asymptotes of a hyperbola are two imaginary lines that the hyperbola is bound by. It can never touch the asymptotes, though it will get very close, just like the definition of asymptotes states.

Asymptotes of a hyperbola are the lines that pass through the center of the hyperbola. The hyperbola gets closer and closer to the asymptotes, but can never reach them.
We know the general equation of a hyperbola.

So comparing the general equation of the hyperbola with $\dfrac{{{x}^{2}}}{25}-\dfrac{{{y}^{2}}}{16}=1$.
We get $a=\pm 5$ and $y=\pm 4$.
So we get the two asymptotes i.e.
So now let’s find out the value of$\dfrac{b}{a}$.
So$\dfrac{b}{a}=\dfrac{\pm 4}{\pm 5}$
So now taking$y=\dfrac{b}{a}x$, we get,
And for$y=-\dfrac{b}{a}x$, we get,
So we get two slopes i.e. ${{m}_{1}}=\dfrac{4}{5}$and ${{m}_{2}}=-\dfrac{4}{5}$.
So now angle between asymptotes$=\tan \theta =\dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}$.
So now substituting the values we get,

$\tan \theta =\dfrac{\dfrac{4}{5}-\left( -\dfrac{4}{5} \right)}{1+\dfrac{4}{5}\left( -\dfrac{4}{5} \right)}=\dfrac{2\left( \dfrac{4}{5} \right)}{1-{{\left( \dfrac{4}{5} \right)}^{2}}}$
Now let $\tan x=\dfrac{4}{5}$.
So substituting$\tan x=\dfrac{4}{5}$above we get,
$\tan \theta =\dfrac{2\tan x}{1-{{\left( \tan x \right)}^{2}}}$
We know the identity \[\tan 2x=\dfrac{2\tan x}{1-{{\left( \tan x \right)}^{2}}}\],
So we get,
$\tan \theta =\tan 2x$
So simplifying in a simple manner we get,
$\theta =2x$
So we know$\tan x=\dfrac{4}{5}$,
So$x={{\tan }^{-1}}\left( \dfrac{4}{5} \right)$
So$\theta =2x=2{{\tan }^{-1}}\left( \dfrac{4}{5} \right)$
So we get the final angle between asymptotes as $2{{\tan }^{-1}}\left( \dfrac{4}{5} \right)$.
So the correct answer is option(B).

Note: Read the question carefully. Don’t confuse yourself in substitution. Here you should be familiar with $\tan \theta =\dfrac{2\tan x}{1-{{\left( \tan x \right)}^{2}}}$and$\tan \theta =\dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}$. Don’t jumble yourself while solving. The mistakes occur while substituting the values in$\tan \theta =\dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}$i.e. of a minus sign. So take care that you do not miss the minus sign.