Question

Show that angle between the two asymptotes of a hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ is $2{\tan ^{ - 1}}\dfrac{b}{a}$ or $2{\sec ^{ - 1}}(e)$.

Answer 1

Hint: Here, we have to show that the angle between the two asymptotes of a hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ is $2{\tan ^{ - 1}}\dfrac{b}{a}$ or $2{\sec ^{ - 1}}(e)$. So, firstly we will calculate the asymptotes of the hyperbola which is given by $y = \pm \dfrac{b}{a}x$ and then we find their slope to calculate the angle between two asymptotes which is given by the formula $\tan \theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}$.

Complete step by step answer:
The hyperbola is one among the three forms of conic sections which is formed by the intersection of a plane and a double cone and can be defined as a type of smooth curve lying in a plane and defined by its geometric properties or by equations for which it is the solution set. 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 and these are the imaginary lines that a function will get very close to, but never touch.

So, the asymptotes of the hyperbola are two imaginary lines that the hyperbola is bound by and it can never touch the asymptote, though it will get very close, as the definition of asymptotes states.Here, we have the equation of hyperbola i.e.,
$\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$
So, the asymptotes of the hyperbola will be $y = \dfrac{b}{a}x$ and $y = - \dfrac{b}{a}x$
Now, we will find the slope of these asymptotes by comparing them with the general equation of a straight line i.e., $y = mx + c$ where $m = $slope of the line.
On comparing we get,
${m_1} = \dfrac{b}{a}$ and ${m_2} = - \dfrac{b}{a}$

Now, we will calculate the angle between two asymptotes using the formula $\tan \theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}$. So,
$ \Rightarrow \tan \theta = \dfrac{{\dfrac{b}{a} - \left( { - \dfrac{b}{a}} \right)}}{{1 + \dfrac{b}{a}\left( { - \dfrac{b}{a}} \right)}}$
On simplifying we get,
$ \Rightarrow \tan \theta = \dfrac{{\dfrac{b}{a} + \dfrac{b}{a}}}{{1 - \dfrac{{{b^2}}}{{{a^2}}}}}$
On further solving, we get,
$ \Rightarrow \tan \theta = \dfrac{{\dfrac{{2b}}{a}}}{{1 - \dfrac{{{b^2}}}{{{a^2}}}}}$

Now let us assume $\tan x = \dfrac{b}{a}$. So, the above equation can be written as
$ \Rightarrow \tan \theta = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} \ldots \ldots (1)$
Now we know that $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
So, we can write the equation $(1)$ as
$ \Rightarrow \tan \theta = \tan 2x$
Cancelling out $\tan $ from the above equation. We get
$ \Rightarrow \theta = 2x$
Now substitute the value $x = {\tan ^{ - 1}}\dfrac{b}{a}$ in the above equation. We get,
$ \Rightarrow \theta = 2\,{\tan ^{ - 1}}\dfrac{b}{a}$
We can also write the above equation as $\theta = 2\,{\sec ^{ - 1}}(e)$

Therefore, the angle between the two asymptotes of a hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ is $2\,{\tan ^{ - 1}}\dfrac{b}{a}$ or $2\,{\sec ^{ - 1}}(e)$.

Note: In order to solve these types of problems the students must be familiar with the formulas such as the general equation of the line i.e., $y = mx + c$ and $\tan \theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}$. Don’t confuse yourself in substituting the values in $\tan \theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}$ there is a chance of mistake of a minus sign while substituting the values. If you miss the negative sign the result may get differ.